Hi,
This my first post on this forum, after reading it for a year or so. I have hard water, and have always been struggling with all forms of algues since starting the hobby.
I am re-doing a tank. I have invested in an RO unit. I am still trying to decide how to proceed next:
1) easy solution to cut my tap water with 50/50 RO. Or more depending on my water supply.
2) or spend more money on remineralizer to have full control.
Green aqua had a recent video on water parameter, advising a GH of 5-6dH, a KH of 2-6dKH, while maintaining GH-2>=KH.
I do understand it is subjective, and personal. But they do have very nice aquascapes.
I have GH, and KH test kit, but I am colorblind so it is a struggle. I found the water report form my water suppliers. I have been trying to decipher it to get to GH, and KH, with the assumption that KH<GH for most tap water in the UK.
Like many, I am well confuse between GH, KH, alkalinity, total hardness. Especially, alkalinity and total hardness which, in my water report, is in both case a measure of concentration of CaCO3. How can you have a different name if it is a measure of the same compound?
I enclose the water report. Looking at mean values, I have an alkalinity of 283mg/l (second entry in the table), and a total hardness (GH?) of 225.81mg/l (quoted at the bottom).

So I turned to Chatgpt, and pasted it the water report. I am amazed at what it came up. I will share the transcript.
It came up with GH=11.6dH, and a KH=15.3dKH. and after cutting it with ~50/50 RO it would be 6dH, and 7.9dKH.
After reading so many post of the subject on the topic, I think I would agree with "him/her/it" 🙂.
Base on what I read online, I was reading my water report with the expectation KH<GH, as GH=KH+NKH so I am well confused. Does it seems correct then?
I am going to go with that, and see what happen. I won't meet the GH-2>KH. What is the drawback to have KH>GH?
I will use CO2 injection. The tank is for a classic planted with some tetra/rasbora and ammano/cherry shrimp.
Also I thought I could share my experience (some other may want to try too) and share the whole transcript below. Hopefully I am not breaking any rules.
The interesting bit, Chatgpt used Ca, and Mg concentration to workout GH.
=================================================================================================
Can you tell me the GH, and KH of my water supply based on the water report below?
From the water quality report you shared:
GH (°dH)=(mg/l of Ca7.14)+(mg/l of Mg4.34)\text{GH (°dH)} = \left(\frac{\text{mg/l of Ca}}{7.14}\right) + \left(\frac{\text{mg/l of Mg}}{4.34}\right)GH (°dH)=(7.14mg/l of Ca)+(4.34mg/l of Mg)
GH (°dH)=83.087.14+3.34.34≈11.64 °dH\text{GH (°dH)} = \frac{83.08}{7.14} + \frac{3.3}{4.34} \approx 11.64 \, \text{°dH}GH (°dH)=7.1483.08+4.343.3≈11.64°dH
So, your General Hardness (GH) is approximately 11.6°dH, which is considered moderately hard.
KH (°dH)=Alkalinity (mg/l)17.848\text{KH (°dH)} = \frac{\text{Alkalinity (mg/l)}}{17.848}KH (°dH)=17.848Alkalinity (mg/l)
KH (°dH)=273.517.848≈15.33 °dH\text{KH (°dH)} = \frac{273.5}{17.848} \approx 15.33 \, \text{°dH}KH (°dH)=17.848273.5≈15.33°dH
Your Carbonate Hardness (KH) is approximately 15.3°dH.
The terms °dH (degree of hardness) and °dKH (degree of carbonate hardness) are both units used to measure aspects of water hardness, but they describe different properties of the water. Here's how they differ:
4o
To convert alkalinity from mg/L (ppm) to °dKH, you can use the formula:
°dKH=Alkalinity (mg/L)17.848\text{°dKH} = \frac{\text{Alkalinity (mg/L)}}{17.848}°dKH=17.848Alkalinity (mg/L)
For an alkalinity of 273.5 mg/L:
°dKH=273.517.848≈15.33 °dKH\text{°dKH} = \frac{273.5}{17.848} \approx 15.33 \, \text{°dKH}°dKH=17.848273.5≈15.33°dKH
4o
Total Hardness refers to the measure of the concentration of calcium (Ca²⁺) and magnesium (Mg²⁺) ions in water. It is an important parameter in water chemistry because these minerals contribute to the "hardness" of water, which affects its behavior in industrial, domestic, and biological systems.
Total Hardness (mg/L as CaCO₃)=2.5×[Ca (mg/L)]+4.1×[Mg (mg/L)]\text{Total Hardness (mg/L as CaCO₃)} = 2.5 \times [\text{Ca (mg/L)}] + 4.1 \times [\text{Mg (mg/L)}]Total Hardness (mg/L as CaCO₃)=2.5×[Ca (mg/L)]+4.1×[Mg (mg/L)]
1 °dH=10 mg/L of CaO1 \, \text{°dH} = 10 \, \text{mg/L of CaO}1°dH=10mg/L of CaO
The formula to convert Total Hardness (in mg/L of CaCO₃) to °dH is:
Total Hardness (°dH)=Total Hardness (mg/L as CaCO₃)17.848\text{Total Hardness (°dH)} = \frac{\text{Total Hardness (mg/L as CaCO₃)}}{17.848}Total Hardness (°dH)=17.848Total Hardness (mg/L as CaCO₃)
Total Hardness=(2.5×83.08)+(4.1×3.3)=207.69 mg/L as CaCO₃\text{Total Hardness} = (2.5 \times 83.08) + (4.1 \times 3.3) = 207.69 \, \text{mg/L as CaCO₃}Total Hardness=(2.5×83.08)+(4.1×3.3)=207.69mg/L as CaCO₃
Convert to °dH:
Total Hardness (°dH)=207.6917.848≈11.64 °dH\text{Total Hardness (°dH)} = \frac{207.69}{17.848} \approx 11.64 \, \text{°dH}Total Hardness (°dH)=17.848207.69≈11.64°dH
This matches the values from your water report, which classifies your water as moderately hard.
4o
It’s a common assumption that KH (carbonate hardness) is usually less than or equal to GH (general hardness), but there are situations where KH can exceed GH due to the distinct chemistry of the water supply. Let me explain how this happens.
4o
To calculate the mixing ratio of your tap water with reverse osmosis (RO) water to achieve a target General Hardness (GH), we can use the dilution formula:
Cm=(Ct×Vt)VmC_m = \frac{(C_t \times V_t)}{V_m}Cm=Vm(Ct×Vt)
Where:
Cm=Ct×(VtVt+VRO)C_m = C_t \times \left(\frac{V_t}{V_t + V_{RO}}\right)Cm=Ct×(Vt+VROVt)
We can solve for the ratio of Vt:VROV_t : V_{RO}Vt:VRO (tap water to RO water):
VtVm=CmCt\frac{V_t}{V_m} = \frac{C_m}{C_t}VmVt=CtCm VtVt+VRO=611.6\frac{V_t}{V_t + V_{RO}} = \frac{6}{11.6}Vt+VROVt=11.66 VtVt+VRO≈0.517\frac{V_t}{V_t + V_{RO}} \approx 0.517Vt+VROVt≈0.517
This means 51.7% of the total volume should come from tap water. The remaining 48.3% will be RO water.
Tap Water : RO Water=0.517🙁1−0.517)≈1:0.93\text{Tap Water : RO Water} = 0.517 : (1 - 0.517) \approx 1 : 0.93Tap Water : RO Water=0.517:(1−0.517)≈1:0.93
For simplicity, you can round this to approximately:
Tap Water : RO Water=1:1\text{Tap Water : RO Water} = 1 : 1Tap Water : RO Water=1:1
4o
To calculate the KH of the mixed water (tap + RO), we use the same proportional mixing formula as for GH:
KHmix=(KHtap×VtVm)+(KHRO×VROVm)\text{KH}{\text{mix}} = (\text{KH}{\text{tap}} \times \frac{V_t}{V_m}) + (\text{KH}{\text{RO}} \times \frac{V{RO}}{V_m})KHmix=(KHtap×VmVt)+(KHRO×VmVRO)
Where:
This means the water will have moderate carbonate hardness and a good buffering capacity, which should help stabilize the pH.
This my first post on this forum, after reading it for a year or so. I have hard water, and have always been struggling with all forms of algues since starting the hobby.
I am re-doing a tank. I have invested in an RO unit. I am still trying to decide how to proceed next:
1) easy solution to cut my tap water with 50/50 RO. Or more depending on my water supply.
2) or spend more money on remineralizer to have full control.
Green aqua had a recent video on water parameter, advising a GH of 5-6dH, a KH of 2-6dKH, while maintaining GH-2>=KH.
I do understand it is subjective, and personal. But they do have very nice aquascapes.
I have GH, and KH test kit, but I am colorblind so it is a struggle. I found the water report form my water suppliers. I have been trying to decipher it to get to GH, and KH, with the assumption that KH<GH for most tap water in the UK.
Like many, I am well confuse between GH, KH, alkalinity, total hardness. Especially, alkalinity and total hardness which, in my water report, is in both case a measure of concentration of CaCO3. How can you have a different name if it is a measure of the same compound?
I enclose the water report. Looking at mean values, I have an alkalinity of 283mg/l (second entry in the table), and a total hardness (GH?) of 225.81mg/l (quoted at the bottom).

So I turned to Chatgpt, and pasted it the water report. I am amazed at what it came up. I will share the transcript.
It came up with GH=11.6dH, and a KH=15.3dKH. and after cutting it with ~50/50 RO it would be 6dH, and 7.9dKH.
After reading so many post of the subject on the topic, I think I would agree with "him/her/it" 🙂.
Base on what I read online, I was reading my water report with the expectation KH<GH, as GH=KH+NKH so I am well confused. Does it seems correct then?
I am going to go with that, and see what happen. I won't meet the GH-2>KH. What is the drawback to have KH>GH?
I will use CO2 injection. The tank is for a classic planted with some tetra/rasbora and ammano/cherry shrimp.
Also I thought I could share my experience (some other may want to try too) and share the whole transcript below. Hopefully I am not breaking any rules.
The interesting bit, Chatgpt used Ca, and Mg concentration to workout GH.
=================================================================================================
Can you tell me the GH, and KH of my water supply based on the water report below?
ChatGPT said:
ChatGPTFrom the water quality report you shared:
General Hardness (GH)
GH measures the total concentration of calcium and magnesium ions in water.- Calcium (Ca): 83.08 mg/l
- Magnesium (Mg): 3.3 mg/l
GH (°dH)=(mg/l of Ca7.14)+(mg/l of Mg4.34)\text{GH (°dH)} = \left(\frac{\text{mg/l of Ca}}{7.14}\right) + \left(\frac{\text{mg/l of Mg}}{4.34}\right)GH (°dH)=(7.14mg/l of Ca)+(4.34mg/l of Mg)
GH (°dH)=83.087.14+3.34.34≈11.64 °dH\text{GH (°dH)} = \frac{83.08}{7.14} + \frac{3.3}{4.34} \approx 11.64 \, \text{°dH}GH (°dH)=7.1483.08+4.343.3≈11.64°dH
So, your General Hardness (GH) is approximately 11.6°dH, which is considered moderately hard.
Carbonate Hardness (KH)
KH is related to alkalinity, often expressed as the concentration of carbonate and bicarbonate ions. The report lists:- Alkalinity: 273.5 mg/l
KH (°dH)=Alkalinity (mg/l)17.848\text{KH (°dH)} = \frac{\text{Alkalinity (mg/l)}}{17.848}KH (°dH)=17.848Alkalinity (mg/l)
KH (°dH)=273.517.848≈15.33 °dH\text{KH (°dH)} = \frac{273.5}{17.848} \approx 15.33 \, \text{°dH}KH (°dH)=17.848273.5≈15.33°dH
Your Carbonate Hardness (KH) is approximately 15.3°dH.
Summary
- GH: ~11.6°dH (Moderately Hard)
- KH: ~15.3°dH (High Alkalinity)
You said:
So what is the difference between dH, and dKH?ChatGPT said:
ChatGPTThe terms °dH (degree of hardness) and °dKH (degree of carbonate hardness) are both units used to measure aspects of water hardness, but they describe different properties of the water. Here's how they differ:
1. °dH (General Hardness)
- What it Measures:
°dH refers to General Hardness (GH), which measures the total concentration of calcium (Ca²⁺) and magnesium (Mg²⁺) ions in the water. These are the minerals responsible for forming limescale in appliances like kettles and washing machines. - Unit Conversion:
1 °dH = 10 mg/L (ppm) of calcium oxide (CaO). It can also be expressed as a combination of calcium and magnesium concentrations. - Relevance:
GH affects fish and plant health in aquariums and indicates the "hardness" of water when considering its effect on household equipment.
2. °dKH (Carbonate Hardness)
- What it Measures:
°dKH refers to Carbonate Hardness (KH), which measures the concentration of carbonate (CO₃²⁻) and bicarbonate (HCO₃⁻) ions. These compounds act as buffers to stabilize pH by neutralizing acids. - Unit Conversion:
1 °dKH = 17.848 mg/L (ppm) of calcium carbonate (CaCO₃). - Relevance:
KH plays a critical role in maintaining stable pH levels in water. Low KH can cause pH swings, which can harm aquatic life, while high KH makes water more resistant to pH changes.
Key Differences
Aspect | °dH (GH) | °dKH (KH) |
---|---|---|
What it Measures | Total calcium and magnesium | Carbonates and bicarbonates |
Effect on Water | Determines "hardness" of water | Buffers pH stability |
Impact | Affects limescale and mineral levels | Affects pH stability and buffering capacity |
Units | °dH = 10 mg/L CaO | °dKH = 17.848 mg/L CaCO₃ |
Summary
- GH (°dH): Measures calcium and magnesium—key for "hard" water.
- KH (°dKH): Measures carbonates and bicarbonates—important for pH buffering.
4o
You said:
An alkalinity of 273.5 mg/l is how much dKH?ChatGPT said:
ChatGPTTo convert alkalinity from mg/L (ppm) to °dKH, you can use the formula:
°dKH=Alkalinity (mg/L)17.848\text{°dKH} = \frac{\text{Alkalinity (mg/L)}}{17.848}°dKH=17.848Alkalinity (mg/L)
For an alkalinity of 273.5 mg/L:
°dKH=273.517.848≈15.33 °dKH\text{°dKH} = \frac{273.5}{17.848} \approx 15.33 \, \text{°dKH}°dKH=17.848273.5≈15.33°dKH
Final Answer:
An alkalinity of 273.5 mg/L is approximately 15.3 °dKH.4o
You said:
What is total hardness?ChatGPT said:
ChatGPTTotal Hardness refers to the measure of the concentration of calcium (Ca²⁺) and magnesium (Mg²⁺) ions in water. It is an important parameter in water chemistry because these minerals contribute to the "hardness" of water, which affects its behavior in industrial, domestic, and biological systems.
Components of Total Hardness
- Calcium Hardness (Ca²⁺):
- The portion of hardness caused by calcium ions in the water.
- Measured in mg/L (milligrams per liter) or ppm (parts per million).
- Magnesium Hardness (Mg²⁺):
- The portion of hardness caused by magnesium ions in the water.
- Measured in mg/L or ppm.
Formula for Total Hardness (in mg/L as CaCO₃):
To calculate total hardness as mg/L of calcium carbonate (CaCO₃), use the formula:Total Hardness (mg/L as CaCO₃)=2.5×[Ca (mg/L)]+4.1×[Mg (mg/L)]\text{Total Hardness (mg/L as CaCO₃)} = 2.5 \times [\text{Ca (mg/L)}] + 4.1 \times [\text{Mg (mg/L)}]Total Hardness (mg/L as CaCO₃)=2.5×[Ca (mg/L)]+4.1×[Mg (mg/L)]
Total Hardness in °dH
Hardness is often expressed in degrees of German Hardness (°dH), where:1 °dH=10 mg/L of CaO1 \, \text{°dH} = 10 \, \text{mg/L of CaO}1°dH=10mg/L of CaO
The formula to convert Total Hardness (in mg/L of CaCO₃) to °dH is:
Total Hardness (°dH)=Total Hardness (mg/L as CaCO₃)17.848\text{Total Hardness (°dH)} = \frac{\text{Total Hardness (mg/L as CaCO₃)}}{17.848}Total Hardness (°dH)=17.848Total Hardness (mg/L as CaCO₃)
Classification of Hardness
Hardness (mg/L as CaCO₃) | Hardness (°dH) | Classification |
---|---|---|
0–60 | 0–3.4 | Soft |
61–120 | 3.4–6.7 | Moderately hard |
121–180 | 6.8–10.1 | Hard |
>180 | >10.1 | Very hard |
Example from Your Report:
Your water report lists:- Calcium (Ca): 83.08 mg/L
- Magnesium (Mg): 3.3 mg/L
Total Hardness=(2.5×83.08)+(4.1×3.3)=207.69 mg/L as CaCO₃\text{Total Hardness} = (2.5 \times 83.08) + (4.1 \times 3.3) = 207.69 \, \text{mg/L as CaCO₃}Total Hardness=(2.5×83.08)+(4.1×3.3)=207.69mg/L as CaCO₃
Convert to °dH:
Total Hardness (°dH)=207.6917.848≈11.64 °dH\text{Total Hardness (°dH)} = \frac{207.69}{17.848} \approx 11.64 \, \text{°dH}Total Hardness (°dH)=17.848207.69≈11.64°dH
This matches the values from your water report, which classifies your water as moderately hard.
4o
You said:
I thought KH had to be less than GH?ChatGPT said:
ChatGPTIt’s a common assumption that KH (carbonate hardness) is usually less than or equal to GH (general hardness), but there are situations where KH can exceed GH due to the distinct chemistry of the water supply. Let me explain how this happens.
Why KH is Sometimes Greater Than GH
- KH and GH Measure Different Things
- KH (carbonate hardness): Represents the concentration of carbonates (CO₃²⁻) and bicarbonates (HCO₃⁻), which act as pH buffers. KH doesn’t directly measure calcium or magnesium.
- GH (general hardness): Measures the total amount of calcium (Ca²⁺) and magnesium (Mg²⁺) ions, which are responsible for "hard water."
- Non-Carbonate Sources of Alkalinity
- KH is influenced by carbonates and bicarbonates, which may be present in higher concentrations due to water treatment processes or natural geology (e.g., limestone-rich areas or artificial buffering agents).
- GH is specifically tied to calcium and magnesium levels, which might be lower in certain water supplies.
- Water Treatment and Additives
- In treated water, carbonate and bicarbonate compounds may be added to stabilize pH or prevent pipe corrosion. This artificially increases KH without affecting GH.
- Softening processes (e.g., ion exchange) may remove calcium and magnesium, reducing GH, while leaving KH largely unchanged.
- Natural Geological Variability
- In some water sources, natural conditions may result in high carbonate content (raising KH) and relatively low calcium/magnesium concentrations (lowering GH).
Specific Case: Your Water Report
- KH: 273.5 mg/L alkalinity = 15.3 °dKH
- GH: ~11.6 °dH
- Your water likely has a high concentration of carbonates and bicarbonates relative to calcium and magnesium.
- This could result from natural geology (e.g., limestone deposits) or water treatment practices.
Typical Scenarios
- KH < GH (Most Common):
- Typical in untreated or naturally hard water with high calcium and magnesium levels.
- KH ≈ GH:
- Occurs when both carbonate and calcium/magnesium levels are balanced, often in natural water sources.
- KH > GH (Your Case):
- Common in treated water or water from regions with carbonate-rich geology and relatively lower calcium/magnesium concentrations.
Summary
Although KH < GH is typical, it's not a rule. Your water has high KH relative to GH due to elevated bicarbonates/carbonates, likely from geological factors or water treatment. Both values are independent, so it’s entirely normal for KH to exceed GH in certain circumstances.4o
You said:
In what proportion to mix my tap water with RO water to achieve an overall GH of 6dH?ChatGPT said:
ChatGPTTo calculate the mixing ratio of your tap water with reverse osmosis (RO) water to achieve a target General Hardness (GH), we can use the dilution formula:
Cm=(Ct×Vt)VmC_m = \frac{(C_t \times V_t)}{V_m}Cm=Vm(Ct×Vt)
Where:
- CmC_mCm: Target GH (desired final hardness, 6 °dH in your case)
- CtC_tCt: Tap water GH (11.6 °dH from your report)
- CROC_{RO}CRO: RO water GH (0 °dH, assuming pure RO water)
- VtV_tVt: Volume of tap water
- VROV_{RO}VRO: Volume of RO water
- Vm=Vt+VROV_m = V_t + V_{RO}Vm=Vt+VRO: Total volume after mixing
Cm=Ct×(VtVt+VRO)C_m = C_t \times \left(\frac{V_t}{V_t + V_{RO}}\right)Cm=Ct×(Vt+VROVt)
We can solve for the ratio of Vt:VROV_t : V_{RO}Vt:VRO (tap water to RO water):
VtVm=CmCt\frac{V_t}{V_m} = \frac{C_m}{C_t}VmVt=CtCm VtVt+VRO=611.6\frac{V_t}{V_t + V_{RO}} = \frac{6}{11.6}Vt+VROVt=11.66 VtVt+VRO≈0.517\frac{V_t}{V_t + V_{RO}} \approx 0.517Vt+VROVt≈0.517
This means 51.7% of the total volume should come from tap water. The remaining 48.3% will be RO water.
Mixing Ratio
The ratio of tap water to RO water is:Tap Water : RO Water=0.517🙁1−0.517)≈1:0.93\text{Tap Water : RO Water} = 0.517 : (1 - 0.517) \approx 1 : 0.93Tap Water : RO Water=0.517:(1−0.517)≈1:0.93
For simplicity, you can round this to approximately:
Tap Water : RO Water=1:1\text{Tap Water : RO Water} = 1 : 1Tap Water : RO Water=1:1
Final Steps:
- Mix equal parts tap water and RO water to achieve a GH close to 6 °dH.
- For higher accuracy, adjust slightly if needed based on testing.
4o
You said:
What would be the KH of such a mix?ChatGPT said:
ChatGPTTo calculate the KH of the mixed water (tap + RO), we use the same proportional mixing formula as for GH:
KHmix=(KHtap×VtVm)+(KHRO×VROVm)\text{KH}{\text{mix}} = (\text{KH}{\text{tap}} \times \frac{V_t}{V_m}) + (\text{KH}{\text{RO}} \times \frac{V{RO}}{V_m})KHmix=(KHtap×VmVt)+(KHRO×VmVRO)
Where:
- KHtap=15.3 °dKH\text{KH}_{\text{tap}} = 15.3 \, \text{°dKH}KHtap=15.3°dKH (from your water report)
- KHRO=0 °dKH\text{KH}_{\text{RO}} = 0 \, \text{°dKH}KHRO=0°dKH (RO water has no carbonate hardness)
- VtVm=0.517\frac{V_t}{V_m} = 0.517VmVt=0.517 (proportion of tap water in the mix)
- VROVm=0.483\frac{V_{RO}}{V_m} = 0.483VmVRO=0.483 (proportion of RO water in the mix)
Calculation:
KHmix=(15.3×0.517)+(0×0.483)\text{KH}{\text{mix}} = (15.3 \times 0.517) + (0 \times 0.483)KHmix=(15.3×0.517)+(0×0.483) KHmix=7.91 °dKH\text{KH}{\text{mix}} = 7.91 \, \text{°dKH}KHmix=7.91°dKHFinal Answer:
The KH of the mixed water would be approximately 7.9 °dKH.This means the water will have moderate carbonate hardness and a good buffering capacity, which should help stabilize the pH.