Question

Diborane, $B_2{H}_6$ reacts with water to form boric acid and hydrogen. what is the pH of the solution which results when 1.104g of $B_2{H}_6$ reacts with 100 $mL$ water?
$If{k}_{a}of{H}_{3}B{O}_3=8\times {10}^{-10},p{K}_a=9.1;AtomicweightofB=10.8g;Molarmassof{B}_2{H}_6=27.6gmo{l}^{-1},log0.8=-0.1$

• A. 4.6
• B. 5.6
• C. 6
• D. 3.12

Answer 1

The correct option is A 4.6

$B_2{H}_6+6H_{2}O\to 2H_{3}B{O}_3+2H_2$
From the above equation 1 mol (27.6 g) of $B_2{H}_6$ gives 2 moles of $H_{3}B{O}_3$
1.104 g of $B_2{H}_6$ gives =$\frac{2\times 1.104}{27.6}=0.08mol$
Since, 0.08 mol of $H_{3}B{O}_3$ is present in 100 $mL$ resulting molarity of $H_{3}B{O}_3$ solution is
$=\frac{0.08mol\times 1000}{100mL}=0.8M$
We know that $H_{3}B{O}_3$ is a weak acid , pH of weak acid is
$pH=\frac{1}{2}(p{K}_a-logC)$
$=\frac{1}{2}(9.1-log0.8)$
$=\frac{1}{2}[9.1-(-0.1\left)\right]$
pH of $H_{3}B{O}_3$ =4.6