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)

greater than 7

Answer 2

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$_2SO_4 = 10 \times 0.025$
\hspace40mm =0.25 \, millimole
$H_2SO_4 +2NaOH \rightarrow Na_2SO_4 +H_2O$
$1mol \, \, \, \, \, \, \, \, 2 \, mol$
0.25 millimole of H$_2SO_4$ 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= $\frac{0.5}{20} =2.5 \times 10^{-2} M$
$\, \, \, \, \, pOH =-log[OH^-]=-log(2.5 \times 10^{-2})$
\hspace30mm =1.60
\hspace30mm pH=12.4