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Chapter 14: Problem 149

A \(10.0\) -mL sample of an \(\mathrm{HCl}\) solution has a \(\mathrm{pH}\) of \(2.000\). What volume of water must be added to change the \(\mathrm{pH}\) to \(4.000\) ?

Short Answer

Expert verified
To change the pH of the HCl solution from 2.000 to 4.000, you need to add \(990\) mL of water.

Step by step solution

01

Find the initial concentration of hydronium ions (H3O+)

To find the initial concentration of hydronium ions, we can use the formula: pH = -log(H3O+) In this case, the pH is 2.0. Rearranging the formula to solve for the H3O+ concentration, we get: H3O+ = 10^(-pH) Plugging in the value of pH = 2.0, we get the initial concentration of hydronium ions: H3O+ = 10^(-2) = 0.01 M
02

Find the final concentration of hydronium ions (H3O+)

Using the same formula as in Step 1, we can find the final concentration of hydronium ions when the pH is changed to 4.0: pH = -log(H3O+) Rearranging the formula to find the H3O+ concentration, we get: H3O+ = 10^(-pH) Plugging in the value of pH = 4.0, we get the final concentration of hydronium ions: H3O+ = 10^(-4) = 0.0001 M
03

Use the dilution formula to find the final volume of the solution

In order to calculate the final volume after adding water, we can use the dilution formula: C1V1 = C2V2 Where C1 and V1 are the initial concentration and volume of the HCl solution, and C2 and V2 are the final concentration and volume of the solution. We already found the initial and final concentrations of hydronium ions in Steps 1 and 2. The initial volume of the HCl solution is given as 10.0 mL. We can rearrange the formula to solve for the final volume V2: V2 = (C1V1) / C2 Plugging in the values, we have: V2 = (0.01 M x 10.0 mL) / 0.0001 M V2 = 1000 mL
04

Find the volume of water to be added

Since the initial volume of the HCl solution is 10.0 mL and the final volume after dilution is 1000 mL, the volume of water to be added to change the pH from 2.0 to 4.0 is: Volume of water = Final volume - Initial volume Volume of water = 1000 mL - 10 mL Volume of water = 990 mL So, to change the pH of the HCl solution from 2.0 to 4.0, 990 mL of water must be added.

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