Question

Calculate the mass of HONH₂ required to dissolve in enough water to make 250.0 mL of solution having a pH of 10.00 (K_b = 1.1 × 10^-8).

Short answer

10.67 g of HONH₂ required to make 250.0 mL of solution having a pH of 10.00.

Step-by-step solution

[OH-] Concentration Calculation of HONH2

It has to be calculated as:

$\begin{array}{c}\text{pOH=14-pH}\\ \text{pOH=14-10=4}\\ {\text{[OH}}^{\text{-}}{\text{]=\hspace{0.17em}10}}^{\text{-pH}}{\text{=10}}^{\text{-4}}\end{array}$

Molarity Calculation of HONH2

Reaction of HONH₂ with water:

$\begin{array}{l}{\text{HONH}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {{\text{\hspace{0.17em}\hspace{1em}HONH}}_3}^{\text{+}}{\text{+\hspace{0.17em}OH}}^{\text{-}}\\ \text{a\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{1em}}-00\\ \text{a}-\text{x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{1em}}-\text{\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{1em}\hspace{1em}x\hspace{1em}\hspace{1em}\hspace{1em}x}(1\times {10}^{-4})\end{array}$

Calculation of HONH₂ concentration using K_b

$\begin{array}{l}{\text{K}}_{\text{b}}=\frac{{\text{x}}^{\text{2}}}{\text{a-x}}=\frac{{(1\times {10}^{-4})}^2}{\text{a}-(1\times {10}^{-4})}=1.1\times {10}^{-8}(Taking,a-(1\times {10}^{-4})~a)\\ \text{a}=\frac{{(1\times {10}^{-4})}^2}{1.1\times {10}^{-8}}=0.909\end{array}$

Mass Calculation of HONH2 

Mole of HONH₂ calculation

$\begin{array}{l}\text{Molarity=}\frac{\text{Mole}}{\text{Volume(L)}}\\ \text{Molarity=}\frac{\text{Mole}}{\text{0.25}}\text{=0.909}\\ \text{Mole=0.909\hspace{0.17em}×\hspace{0.17em}0.25=0.227}\end{array}$

$\begin{array}{l}\text{Mole=}\frac{\text{Mass}}{\text{Molarmass}}\\ \text{Mass=Mole×Molar\hspace{0.17em}mass}\\ \text{Mass=\hspace{0.17em}0.227}\times 47=10.67\text{\hspace{0.17em}g}\end{array}$