How many litres of water must be added to 1 L of an aqueous solution of HCl with a pH of 1 to create

Q: Why do we use the formula $M_1V_1 = M_2V_2$ here?

When you add water to a solution, you are only changing the volume of the solvent. The actual number of acid molecules (moles of solute) remains exactly the same. Since Molarity $\times$ Volume = Moles, the product $M \times V$ before dilution must equal the product $M \times V$ after dilution.

Q: Can I just say that going from pH 1 to pH 2 means adding 1 litre of water?

No, that is a very common trap! The pH scale is logarithmic, not linear. A change of 1 unit in pH actually means the concentration has changed by a factor of 10. Therefore, the volume must also change by a factor of 10.

Q: What if the acid in the question was a weak acid like acetic acid?

If it were a weak acid, you could not apply $M_1V_1 = M_2V_2$ directly to the $[H^+]$ concentration. According to Ostwald's Dilution Law, the degree of dissociation ($\alpha$) of a weak acid increases as you add water. You would have to account for the changing equilibrium.

Q: Why is the final answer 9 L and not 10 L? I calculated $V_2 = 10$.

You correctly calculated that the *total final volume* must be 10 L. However, the question specifically asks 'How many litres of water must be *added*'. Since you already started with 1 L in the beaker, you only need to add 9 L more to reach the 10 L mark.

Q: Does the number of $H^+$ ions change when we add water?

For a strong acid like HCl, it is already 100% dissociated. Adding water does not create more $H^+$ ions; it simply spreads the existing ions out over a larger volume, which is why the concentration (and thus the acidity) decreases.

Question

How many litres of water must be added to 1 L of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

Options:

(A) 0.1 L

(B) 0.9 L

(C) 2.0 L

(D) 9.0 L

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing in a chemistry lab with a beaker containing exactly $1 	ext{ L}$ of a hydrochloric acid (HCl) solution. We are given a crucial piece of information: the initial pH of this solution is exactly $1$. Our objective is to dilute this solution by adding pure water until its pH rises to $2$.

To solve this, we must first translate the abstract concept of pH into a tangible physical quantity: molar concentration. Remember the fundamental definition of pH? It is the negative base-10 logarithm of the hydrogen ion concentration, mathematically expressed as $	ext{pH} = -\log_{10}[H^+]$.

By rearranging this formula, we can easily find the concentration: $[H^+] = 10^{-	ext{pH}}$.

### The Master Equation
Let's apply this to our initial state. Since the initial pH is $1$, the initial molarity of hydrogen ions, which we will call $M_1$, is $10^{-1} 	ext{ M}$, or $0.1 	ext{ M}$.

Now, what about our target state? We want the final pH to be $2$. This means our final molarity, $M_2$, must be $10^{-2} 	ext{ M}$, which is $0.01 	ext{ M}$. Notice how an increase of just $1$ unit on the logarithmic pH scale corresponds to a massive $10$-fold decrease in the actual concentration of acid!

When we add pure water to dilute a solution, the number of moles of the acid solute remains absolutely constant. Only the total volume of the solution changes. This physical reality gives us the powerful dilution equation:

$$M_1V_1 = M_2V_2$$

### Final Calculation
Let's substitute our known values into the dilution equation. We know $M_1 = 0.1 	ext{ M}$, $V_1 = 1 	ext{ L}$, and $M_2 = 0.01 	ext{ M}$.

$$0.1 	imes 1 = 0.01 	imes V_2$$

Solving for $V_2$, we find:

$$V_2 = \frac{0.1}{0.01} = 10 	ext{ L}$$

**There is a catch here!** The question specifically asks for the volume of water *added*, not the final total volume. We already had $1 	ext{ L}$ of solution in the beaker initially. Therefore, the volume of pure water we need to pour in is the difference between the final volume and the initial volume.

$$	ext{Volume added} = V_2 - V_1 = 10 	ext{ L} - 1 	ext{ L} = \mathbf{9 	ext{ L}}$$

Thus, we must add exactly $9 	ext{ L}$ of water to achieve a pH of $2$.

Animated Solution for Chemistry - Ionic Equilibrium: How many litres of water must be added to 1 L of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

Select Answer:

Visualized Solution

$Initial State$

$pH to Concentration$

$Initial Concentration$

$Final Concentration$

$Dilution Law$

$Finding Final Volume$

$Volume of Water Added$

$Food for Thought$

The Sigma Insight: pH, Buffer and Indicator

Analyzing the Setup Imagine you are standing in a chemistry lab with a beaker containing exactly $1 L$ of a hydrochloric acid (HCl) solution

The Master Equation Let's apply this to our initial state

Final Calculation Let's substitute our known values into the dilution equation

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Animated Solution for Chemistry - Ionic Equilibrium: How many litres of water must be added to 1 L of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pH to Concentration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Concentration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Concentration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dilution Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Final Volume

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Volume of Water Added

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought