# What volume of water (in mL) must be added to 238 mL of 0.405 M hydrochloric acid to create a...

## Question:

What volume of water (in mL) must be added to 238 mL of 0.405 M hydrochloric acid to create a solution with a pH of 0.681?

(Give the answer in standard notation.)

## Dilution

The addition of water in the solution is known as the dilution. We can dilute the solution to get the solution of the desired concentration or pH. By diluting the solution, we can decrease the concentration of the ions per unit volume of the solution.

Given:

• Molarity of {eq}HCl {/eq} = 0.405 M
• Volume of {eq}HCl {/eq} = 238 mL
• pH after addition of water = 0.681

We are told to find the volume of water added to the solution and for this we will find the molarity of {eq}H^+ {/eq} with the help of the pH value so that we get the molarity of {eq}HCl {/eq} solution after dilution.

{eq}pH = -log[H^+]\\ [H^+] = 10^{-pH}\\ [H^+] = 10^{-0.681}\\ [H^+] = 0.208\ M {/eq}

As 1 mole of {eq}H^+ {/eq} is released for 1 mole of {eq}HCl {/eq}, hence, the molarity of {eq}HCl {/eq} after dilution is 0.208 M

Now we will find the moles of {eq}HCl {/eq} before dilution as follows:

{eq}Moles = molarity\times volume\\ Moles = 0.405\times0.238\\ Moles = 0.0964\ mol {/eq}

Moles of {eq}HCl {/eq} remain same in the solution after the addition of water. So, now we will find the volume of solution after addition of water as follows:

{eq}Molarity = \dfrac{moles}{volume}\\ 0.208\ M = \dfrac{0.0964\ mol }{Volume} Volume = \dfrac{0.0964\ mol}{0.208\ mol/L}\\ Volume = 0.4634 \ L\\ Volume = 463.4\ mL {/eq}

Now we will find the volume of water added as follows:

Volume of water added = {eq}463.4 - 238 = 225.4\ mL{/eq}

Hence, the volume of water added is equal to {eq}2.25\times10^2\ mL {/eq} Calculating Dilution of Solutions

from

Chapter 8 / Lesson 5
71K

Learn what a solution is and how to properly dilute a new solution from a stock solution. Learn the dilution equation that combines molarity, the volume of stock solution and desired solution to determine how much stock solution is needed for the new solution.