Question

A solution contains $10\, mL\, 0.1 \,N \,NaOH$ and $10 \,mL \,0.05 NH_2SO_4 ,PH$ of this solution is

• A. less than 7
• B. 7
• C. zero
• D. greater than 7

Answer 1

Answer: (D)

$10\, mL$ of $0.1 \,N (0.1 \,M ) NaOH =10 \times 0.1=1$ millimole
$10 \,mL$ of $0.05 \,N (0.025 M ) H _{2} SO _{4}=10 \times 0.025$
$=0.25$ millimole
$H _{2} SO _{4}+2 NaOH \rightarrow Na _{2} SO _{4}+ H _{2} O$
$1 \,mol \,2 \,mol$
$0.25$ millimole of $H _{2} SO _{4}$ will react with
$0.5$ millimole of $NaOH$.
$\therefore NaOH$ left $=1-0.5=0.5$ millimole
Volume of reaction mixture $=20 \,mL$
Molarity of $NaOH$ in mixture solution
$=\frac{0.5}{20}=2.5 \times 10^{-2} M$
$p O H=-\log \left[O H^{-}\right]=-\log \left(2.5 \times 10^{-2}\right)$
$=1.60$
$p H=12.4$