Question

Calculate the $\mathbf{\text{pH}}$of a $\mathbf{\text{1L}}$solution containing

(a) $\mathbf{\text{10mL}}$of$\mathbf{\text{5MNaOH}}$ ,

(b)$\mathbf{\text{10mL}}$ of $\mathbf{\text{100mM}}$glycine and $\mathbf{\text{20mL}}$of $\mathbf{\text{5MHCl}}$, and

(c) $\mathbf{\text{10mL}}$of $\mathbf{\text{2M}}$acetic acid and $\mathbf{\text{5g}}$of sodium acetate (formula weight ${\mathbf{\text{82g}}}^{\mathbf{\text{*}}}{\mathbf{\text{mol}}}^{\mathbf{\text{-1}}}$).

Short answer

1. The pH of a $\text{1L}$ solution is $\text{12.7}$.
2. The pH of a $\text{1L}$ solution is 1 .
3. The pH of a $\text{1L}$solution is $\text{5.24}$.

Step-by-step solution

Definition of Concept

In chemistry, the pH scale, which has historically stood for "potential of hydrogen," is used to describe how acidic or basic an aqueous solution is. The pH values of acidic solutions are typically lower than those of basic or alkaline solutions.

Step 2:Calculate the  pH of a  1 L solution

(a)Consider the given information:

$\text{10mL}$of,$\text{5MNaOH}$

${\text{Kw=10}}^{\text{-14}}$

$\text{10mL=0.01L}$

Calculation for pH,

$\begin{array}{c}\text{pH=}\frac{\text{(0.01L)(5MNaOH)}}{\text{1L}}\\ \text{pH=0.05MNaOH}\\ {\text{=0.05MOH}}^{\text{-}}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\frac{\text{Kw}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \text{=}\frac{{\text{10}}^{\text{-14}}}{\text{0.05M}}\\ {\text{=2×10}}^{\text{-13}}\\ \text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\\ \text{=-log}\left({\text{2×10}}^{\text{-13}}\text{M}\right)\\ \text{ph=12.7}\end{array}$

Therefore, the required pH is $\text{12.7}$.

Calculate the pH of  a 1 L solution

(b)Consider the given information:

$\text{10mL}$of $\text{100mM}$glycine,

$\text{20mL}$of $\text{5MHCl}$.

Calculation for pH,

$\begin{array}{r}\frac{\text{(0.02L)(5MHCl)}}{\text{1L}}\text{=0.1MHCl}\\ {\text{=0.1MH}}^{\text{+}}\\ \text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\\ \text{=-log[0.1M]}\\ \text{pH=1}\end{array}$

For glycine:

$\frac{\text{(0.01L)(100mM)}}{\text{1L}}\text{=1mM}$glycine is insignificant in strength to HCI.

Therefore, the required pH is 1 .

Calculate the pH of  a 1L  solution

(c)Consider the given information:

$\text{10mL}$of $\text{2M}$acetic acid,

5g sodium acetate.

1 Determine the acid and base concentrations:

Acetic acid:

$\frac{\text{(0.01L)(2M)}}{\text{1L}}\text{=0.02M}$

Sodium acetate:

$\frac{\text{(5g)}\left(\frac{\text{1mol}}{\text{82g}}\right)}{\text{1L}}\text{=0.061M}$

2 Apply the Henderson-Hesselbach formula:

$\text{pH=pK+log([}$$\text{acetate}\left]\text{/}\right[\text{aceticacid])}$

acetic acid pK: $\text{4.76}$

$\begin{array}{l}\text{pH=4.76+log(0.061/0.02)}\\ \text{pH=5.24}\end{array}$

Therefore, the required pH is $\text{5.24}$.