Question

$100$ mL of $0.2$ N HCl is added to $100$ mL of $0.18$ N. NaOH and the whole volume is made one litre. The pH of the resulting solution is:

• A. $1.2$
• B. $2.7$
• C. $3.9$
• D. $4$

Answer 1

Answer: (B)

$2.7$

100 ml of 0.2 N HCl

So the moles of $\left[H^{+}\right]$ = $100\times$ $\frac{0.2}{1000}$ = 0.02 moles

moles of $\left[O{H}^{+}\right]$ = $100\times$ $\frac{0.18}{1000}=0.018$ moles

$\left[H^{+}\right]$ = $\frac{molesof{H}^{+}}{1000molofsolution}$

$\left[H^{+}\right]$ = $2\times {10}^{-}$${}^3$

pH = -log $\left[H^{+}\right]$ = - log $2\times {10}^{-}$${}^3$

= - log 2 + 3

= 2. 6989