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

Following are the responses to the given choices:

Step-by-step explanation:

For point a:

Using the acid and base which are strong so,

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

`= 24.9\ mL * 0.100\ M \\\\= (24.9)/(1000\ L) * 0.100\ M \\\\= 2.49 * 10^(-3) \ mol`

moles of
`OH^(-)` (from
`KOH`)

`= 25.0\ mL * 0.100\ M \\\\= (25.0)/(1000 \ L) * 0.100 \ M \\\\\= 2.50 * 10^(-3)\ mol`

`1\ mol H^(+) \ neutralizes\ 1\ mol\ of\ OH^(-)`

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

`p[OH^(-)] = -\log [OH^(-)] = -\log [2 * 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 * 10^(-3) \ mol`

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

`= 25.1 mL * 0.100 M\\\\ = (25.1)/(1000)\ L * 0.100 \ M\\\\ = 2.51* 10^(-3) \ mol`

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

So,
`(2.51 * 10^(-3)\ mol - 2.50 * 10^(-3)\ mol)` i.e.
`1 * 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 * 10^(-4)\ M`

Hence,
`pH = -\log [H^+] = -\log[2 * 10^(-4)] = 3.70`