Question

A 50.0 ml sample of 0.200 M HNO2 (weak acid) is titrated with 0.400 M

KOH. Calculate the pH after addition of 25 ml KOH (Ky for HNO, is 4.6 x
10-4)
O a 6.67
O b. 7.67
O c.8.23
O d. 9.12

Answer 1

pH = 8.23

Further explanation

Reaction

HNO₂+KOH⇒KNO₂+H₂O

mol HNO₂ = 50 x 0.2 = 10 mlmol

mol KOH = 0.4 x 25 = 10 ml mol

Both of them completely react, resulting in hydrolysis salt which is composed of weak acids and strong bases

Formula :

`\tt [OH^-]=\sqrt{(Kw)/(Ka).M }`

Kw = water constant = 10⁻¹⁴

Ka = 4.6 x 10⁻⁴

M = anion (salt) concentration =

`\tt (mol~KNO_2)/(total~volume)=(10)/(50+25)=0.133~M`

`\tt [OH^-]=\sqrt{(10^(14))/(4.6* 10^(-4)) }* 0.133\\\\(OH^-)=1.7* 10^(-6)\\\\pOH=6-log~1.7=5.77\\\\pH+pOH=14\\\\pH+5.77=14\\\\pH=8.23`