Question

If the K a of a monoprotic weak acid is 1.4 × 10 − 6 , what is the pH of a 0.11 M solution of this acid

Answer 1

Assume a weak monoprotic acid is HA. Consider dissociation of weak acid HA in water.

HA (aq) + H₂O (l) H₃O ⁺ (aq) + A ^- (aq)  

Equilibrium constant for above reaction is, K a = [ H₃O ⁺ ] [ A ^- ] / [ HA ] = 1.4 10 ^-06

Let's use ICE table.

Concentration (M)	HA (aq) H3O + (aq) + A - (aq)
I	0.11
C	- X	+X	+X
E	0.11- X	X	X

Substituting these values in K a = [ H₃O ⁺ ] [ A ^- ] / [ HA ] = 1.4 10 ^-06 , we get

( X ) ( X ) / 0.11 - X =1.4 10 ^-06

X ² / 0.11 - X = 1.4 10 ^-06  

X ² = ( 0.11 - X ) 1.4 10 ^-06  

X ² =1.54 10 ^-07 -  1.4 10 ^-06  X

X ² +  1.4 10 ^-06  X - 1.54 10 ^-07 = 0

Comparing above equation with a X ² + b X+ c = 0 , we get a = 1 , b =1.4 10 ^-06 , c = - 1.54 10 ^-07

Now , solve equation for X.

X = - b +/- b ² -4 ac / 2 a  

X = - 1.4 10 ^-06 + /- ( 1.4 10 ^-06 ) ² - 4 (1) ( - 1.54 10 ^-07 ) / 2 (1)

X = - 1.4 10 ^-06 + /- 6.16 10 ^-07 / 2

X = - 1.4 10 ^-06 + /- 7.85 10 ^-04 / 2

X = - 1.4 10 ^-06 + 7.85 10 ^-04 / 2 = 0.000392 M

Or X = - 1.4 10 ^-06 - 7.85 10 ^-04 / 2 = - 0.000393 M

We can accept X = 0.000392 M

We have, X = [ H₃O ⁺ ] = [ A ^- ] = 0.000392 M

We have relation, pH = - log [ H₃O ⁺ ]

pH = - log ( 0.000392 )

pH = 3.41