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Solved: What is the pH of a solution formed by mixing 125.0 mL of 0.0250 M HCl with 75.0

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 109P Chapter 14

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 109P

Problem 109P

What is the pH of a solution formed by mixing 125.0 mL of 0.0250 M HCl with 75.0 mL of 0.0500 M NaOH?

Step-by-Step Solution:

Solution 109P :

Step 1:

Given :

Volume of HCl = 125.0 mL

Molarity of HCl = 0.0250 M

Volume of NaOH = 75.0 mL

Molarity of NaOH = 0.0500 M

pH of solution formed by mixing HCl and NaOH= ?

Here, HCl is a strong acid and NaOH is a strong base.

Let’s write the balanced chemical equation for it :

                HCl(aq) + NaOH(aq) → NaCl + H2O

First, let’s find the number of moles of HCl and NaOH :

We have the molarity and volume of HCl and NaOH, using this we will calculate the number of moles using the molarity formula :

                Molarity =

The formula for calculating the number of moles will be :

Number of moles = Molarity  Volume(in L)   

Volume of HCl and NaOH is in mL, so let’s convert to L :

1 L = 1000 mL

Therefore, 125 mL HCl in L will be :

                = 125 mL

= 0.125 L

75 mL NaOH in L will be :

                = 75 mL

= 0.075 L

Hence, number of moles in HCl is :

                = 0.0250 M 0.125 L

                = 0.003125 moles.

Number of moles in NaOH is :

                =  0.0500 M0.075 L

                = 0.00375 moles.

Now, if we take the difference between [H3O] and [OH] :

 = 0.00375 - 0.003125

 = 0.000625 moles

This tells that there are more [OH] in solution. Hence,  it is a basic solution and there are 0.000625 moles excess in the solution

Step 2 of 3

Chapter 14, Problem 109P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Solved: What is the pH of a solution formed by mixing 125.0 mL of 0.0250 M HCl with 75.0