Question

$40ml,\frac{N}{4}NaOH,20ml\frac{N}{2}HN{O}_3$ and $30ml,\frac{M}{3}HCI$ solutions are mixed and the volume was made upto $1d{m}^3$. Find $pH$ of resulting solution.

• A. $1$
• B. $2$
• C. $8$
• D. $11$

Answer 1

Answer: (B)

$2$

$Numberofmilliequivalents=Molarity\times Volume$

$Molarity=\frac{Normality}{n-factor}$

$MilliequivalentsofNaOH=0.04\times \frac{1}{4}=0.01$

$MilliequivalentsofHN{O}_3=0.02\times \frac{1}{2}=0.01$

$MilliequivalentsofHCl=0.03\times \frac{1}{3}=0.01$

Due to the presence of strong acid and strong base, one of the acid and base ($NaOH$) will undergo neutralisation reaction and only one acid remains in the solution, whose concentration will determine the pH of the solution.

Milli equivalents of acid will remain the same before and after dilution. Let M be the molarity of acid in the resultant 1L of diluted solution.

milli equivalents of acid initially = milli equivalents of acid after dilution

$0.01=M\times 1$

$M=0.01$

Since the acid is monoprotic, so $H^{+}$ concentration of acid $=Molarityofsolution$

$pH=-log\left[H^{+}\right]$

$pH=-log\left[0.01\right]=2$