Question

A solution is made by mixing exactly 500 mL of 0.156 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below. Ka of CH3COOH is 1.8 × 10−5

[H+]

× 10 M
Enter your answer in scientific notation.


[OH−]

M


[CH3COOH]

× 10 M
Enter your answer in scientific notation.


[Na+]

M


[CH3COO−]

M

Answer 1

The hydrogen ion concentration is
`3.6 * 10^(-13)M`.

Step-by-step explanation:

From the given,

Volume of NaOH = 500 ml = 0.500L

Moarity of NaOH = 0.156M

The number of moles of NaOH is calculated is a follows.

`M=(n)/(V)`

Substitute the given values.

`0.156M =(n)/(0.500L)`

`n=0.078\,mole`

Volume of
`CH_(3)COOH` = 500 ml = 0.500L

Moarity of
`CH_(3)COOH` = 0.100M

The number of moles of
`CH_(3)COOH` is calculated is a follows.

`M=(n)/(V)`

Substitute the given values.

`0.100M =(n)/(0.500L)`

`n=0.05\,mole`

NaOH is a strong electrolyte

The number of moles NaOH left in the solution is

`0.078 mol- 0.05=0.028\,mol`

Therefore, hydroxide ion concentration is 0.028 mol

The hydrogen concentration

`[H^(+)][OH^(-)]=10^(-14)`

`[H^(+)]= \frac {10^(-14)}{[OH^(-)]}`

`[H^(+)]= \frac {10^(-14)}{0.028}=3.6 * 10^(-14)`

Therefore, hydrogen ion concentration is
`3.6 * 10^(-14)`,