Question

Calculate pH of mixture containing 50 ml 0.1 (M) NaOH and 50 ml of 0.1 (M) CH3COOH.[Ka = 4.72]

Answer 1

Answer: pH of the mixture is 8.71

Explanation: Reaction is given as:

$CH_3COOH+NaOH\rightleftharpoons H_2O+CH_3COONa$

Reaction of NaOH and $CH_3COOH$ combine in 1:1 ratio to produce $CH_3COONa$

Number of moles can be calculate by

$\text{Number of moles}=Molarity\times Volume(L)$

Molarity of NaOH = 0.1M

Volume of NaOH = 50mL = 0.05L

$\text{Moles of NaOH}= 0.1\times 0.05$

Moles of NaOH = 0.005 moles

Total Volume of the mixture = 100mL = 0.1L

$\text{Moles of }CH_3COONa=0.005moles$

Concentration of $CH_3COONa$ will be

$CH_3COONa=\frac{0.005}{0.1}$ = 0.05moles/L

For salt hydrolysis of weak acid ( $CH_3COOH$ ) and strong base ( NaOH )

$pH=\frac{1}{2}\left[pk_w+pk_a+logC\right]$

where, $pk_w$ = 14

$pk_a$ = 4.72 (given)

C = Concentration of $CH_3COONa$ that is 0.05moles/L

Putting the values in above equation, we get

$pH=\frac{1}{2}\left[14+4.72+log(0.05)]$

pH = 8.71

Answer 2

Answer:

8.71

here's the answer and hope this helps you.