Question

The pH of a solution prepared by mixing 40.0 mL of 0.125 M Mg(OH)2 and 150.0 mL of 0.125 M HCl is ________.

Answer 1

The balanced chemical reaction is:

Mg(OH)2  +  2HCl  -->  MgCl2  +  2H2O

 

So we can see that 2 moles of HCl reacts for every 1 mole of Mg(OH)2.

moles HCl left = 0.125 M * 0.150 L – (0.125 M * 0.040 L) * 2

moles HCl left = 0.00875 mol

 

So the concentration of H+ ion is:

[H+] = 0.00875 mol / (0.150 L + 0.040 L)

[H+] = 0.046 M

 

So the pH is:

pH = - log 0.046

pH = 1.34

Answer 2

The pH of a solution prepared by mixing 40.0 mL of 0.125 M ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ and 150.0 mL of 0.125 M HCl is $\boxed{1.34}$.

Further Explanation:

pH can be used to predict acidity or basicity of substances. Its range varies from 0 to 14. It is defined as negative logarithmof hydrogen ion concentration. Solutions with pH 7 are neutral, those with pH less than 7 are acidic and those having pH more than 7 are alkaline.

The expression for pH is mentioned below.

${\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right]$                                                                     …… (1)

Where $\left[ {{{\text{H}}^ + }} \right]$  is the concentration of hydrogen ion.

Given reaction occurs as follows:

${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}} + 2{\text{HCl}} \to {\text{MgC}}{{\text{l}}_2} + 2{{\text{H}}_{\text{2}}}{\text{O}}$  

This reaction indicates that one mole of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ reacts with two moles of HCl to form one mole of ${\text{MgC}}{{\text{l}}_{\text{2}}}$ along with two moles of ${{\text{H}}_{\text{2}}}{\text{O}}$.

Moles of component can be calculated as follows:

${\text{Moles of component}} = \left( {{\text{Molarity}}} \right)\left( {{\text{Volume}}} \right)$                …… (2)

Substitute 150.0 mL for volume and 0.125 M for molarity in equation (2) to calculate molesof HCl.

$\begin{aligned}{\text{Moles of HCl}} &= \left( {{\text{0}}{\text{.125 M}}} \right)\left( {{\text{150}}{\text{.0 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.01875{\text{ mol}} \\\end{aligned}$  

Substitute 150.0 mL for volume and 0.125 M for molarity in equation (2) to calculate moles of  .

$\begin{aligned}{\text{Moles of Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}} &= \left( {{\text{0}}{\text{.125 M}}} \right)\left( {{\text{40}}{\text{.0 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.005{\text{ mol}} \\\end{aligned}$  

Moles of HCl left after reaction can be calculated as follows:

$\begin{aligned}{\text{Moles of HCl left}} &= 0.01875{\text{ mol}} - 2\left( {0.005{\text{ mol}}} \right) \\&= 0.00875{\text{ mol}} \\\end{aligned}$  

Since HCl is a strong acid, it completely dissociates to produce hydrogen ions. So 0.00875 moles of HCl will be able to form the same amount of hydrogen ions.

Total volume of the solution can be calculated as follows:

 $\begin{aligned}{\text{Volume of solution}} &= \left( {40.0{\text{ mL}} + 150.0{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.19{\text{ L}} \\\end{aligned}$

The concentration of ${{\text{H}}^ + }$ ion can be calculated as follows:

$\begin{aligned} \left[ {{{\text{H}}^ + }} \right] &= \frac{{0.00875{\text{ mol}}}}{{0.19{\text{ L}}}} \\&= 0.046{\text{ M}} \\\end{aligned}$  

Substitute 0.046 M for $\left[ {{{\text{H}}^ + }} \right]$ in equation (1).

$\begin{aligned}{\text{pH}} &= - \log \left( {0.046{\text{ M}}} \right) \\&= 1.34 \\\end{aligned}$  

Learn more:

1. Write the chemical equation responsible for pH of buffer containing  and  : brainly.com/question/8851686
2. Reason for the acidic and basic nature of amino acid. brainly.com/question/5050077

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Acids, base and salts

Keywords: pH, H+, 0.046 M, 1.34, 0.00875 mol, 0.19 L, 40.0 mL, 50.0 mL, HCl, Mg(OH)2, MgCl2, H2O, solution.