Question

g Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH solution.

Answer 1

Answer:

Following are the responses to the given choices:

Explanation:

For point a:

Using the acid and base which are strong so,

moles of $H^+$ (from$HNO_3$)

$= 24.9\ mL \times 0.100\ M \\\\= \frac{24.9}{1000\ L} \times 0.100\ M \\\\= 2.49 \times 10^{-3} \ mol$

moles of $OH^{-}$ (from $KOH$)

$= 25.0\ mL \times 0.100\ M \\\\= \frac{25.0}{1000 \ L} \times 0.100 \ M \\\\\= 2.50 \times 10^{-3}\ mol$  

$1\ mol H^{+} \ neutralizes\ 1\ mol\ of\ OH^{-}$

So,  $(2.50 \times 10^{-3} mol - 2.49 \times 10^{-3} mol)$ i.e. $1 \times 10^{-5}$ mol of $OH^-$ in excess in total volume $(24.9+25.0) \ mL = 49.9 \ mL$ i.e. concentration of $OH^- = 2 \times 10^{-4}\ M$

$p[OH^{-}] = -\log [OH^{-}] = -\log [2 \times 10^{-4}\ mol] = 3.70$

Since, $pH + pOH = 14,$

so,

$\to pH = 14- pOH = 14- 3.70 = 10.30$  

For point b:

moles of $OH^-$ = from point a $= 2.50 \times 10^{-3} \ mol$

moles of $H^+$(from$HNO_3$):

$= 25.1 mL \times 0.100 M\\\\ = \frac{25.1}{1000}\ L \times 0.100 \ M\\\\ = 2.51\times 10^{-3} \ mol$

1 mol $H^+$ neutralizes 1 mol of $OH^-$

So, $(2.51 \times 10^{-3}\ mol - 2.50 \times 10^{-3}\ mol)$ i.e. $1 \times 10^{-5} \ mol \ of\ H^+$ in excess in the total volume of $(25.1+25.0) \ mL = 50.1\ mL$ i.e. concentration of$H^+ = 2 \times 10^{-4}\ M$

Hence, $pH = -\log [H^+] = -\log[2 \times 10^{-4}] = 3.70$