Question

Calculate pH of 0.1M of weak mono basic acid whose dissociation constant is 4×10^-10 @ 298 K

Answer 1

Dear Student,

◆ Answer -

pH = 10.4

● Explanation -

# Given -

Ka = 4×10^-10

[acid] = 0.1 M

T = 298 K

# Solution -

pH of the given monobasic acid is -

pH = pKa + log([salt]/[acid])

pH = -log(4×10^-10) + log(1/0.1)

pH = -(−9.398) + log10

pH = 9.398 + 1

pH = 10.4

Hence, pH of the given monobasic acid is 10.4 .

*This is practically impossible as pH of acid should be below 7. But that's what calculation shows.

Thanks dear. Hope this helps you..

Answer 2

$\mathfrak{Given:-}$

• Ka = 4×10^-10
• [acid] = 0.1 M
• T = 298 K

$\mathfrak{Find:-}$

• Calculate pH of 0.1M of weak mono basic acid

$\mathfrak{Solution:-}$

pH of the given monobasic acid is:-

➜ pH = pKa + log([salt]/[acid])

➜ pH = -log(4×10^-10) + log(1/0.1)

➜ pH = -(−9.398) + log10

➜ pH = 9.398 + 1

➜ pH = 10.4

Hence,

$\textsf {pH of the given monobasic acid is 10.4.}$