Q. A solution contains 10 mL of 0.1 N $NaOH\ and\ 10\ mL\ of\ 0.05\ N\ H_{2}SO_{4}$, pH of this solution is
A) Less than 7
B) 7
C) Zero
D) Greater than 7
Explanation
(d)
Number of milliequivalents of
$NaOH = 10 \times 0.1 = 1$
Number of milliequivalents of
$H_{2}SO_{4} = 10 \times 0.05 = 0.5$
Milliequivalent of $H_{2}SO_{4}\ will\ neutralise\ NaOH$ so, that (1-0.5)=0.5 millieq. Of $NaOH$ is left unneutralised.
Hence, pH of the resulting solution will be greater than seven (pH>7).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.