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Question: 0.1M of HA is titrated with 0.1 M NaOH, calculate the pH at the end point. Given, Ka (HA)=\(5 \times...

0.1M of HA is titrated with 0.1 M NaOH, calculate the pH at the end point. Given, Ka (HA)=5×10−65 \times {10^{ - 6}} and a<<1.

Explanation

Solution

Hint: In this question use the concept of equilibrium Ka=[H+][A−][HA]=cα×cαc(1−α){K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}} = \dfrac{{c\alpha \times c\alpha }}{{c(1 - \alpha )}} to find the value of α\alpha . Then subtract α\alpha from the total number of moles of substance to get the number of ions left un-neutralized. Find the log of the value obtained and subtract it from 14 to get the value of pH at the end point.

Complete answer:
Formula Used:
HA⇌H++A−HA \rightleftharpoons {H^ + } + {A^ - }
Ka=[H+][A−][HA]=cα×cαc(1−α){K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}} = \dfrac{{c\alpha \times c\alpha }}{{c(1 - \alpha )}}
According to question
c = 0.1 M of HA (given)
HA⇌H++A−HA \rightleftharpoons {H^ + } + {A^ - }
At equilibrium
Ka=[H+][A−][HA]=cα×cαc(1−α){K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}} = \dfrac{{c\alpha \times c\alpha }}{{c(1 - \alpha )}}
given a<<1,a < < 1,then(1-a)≈\approx1 can be neglected

5×10−6=cα215 \times {10^{ - 6}} = \dfrac{{c{\alpha ^2}}}{1}
α=0.7071×10−3\alpha = 0.7071 \times {10^{ - 3}}
[H+]=[cα]=0.707×10−4M[{H^ + }] = [c\alpha ] = 0.707 \times {10^{ - 4}}M
0.7071×10−3M0.7071 \times {10^{ - 3}}M will neutralize 0.7071×10−4M0.7071 \times {10^{ - 4}}M of [OH-] of 0.1M NaOH
[OH-] left un-neutralized = 0.1M-0.7071×10−4M0.7071 \times {10^{ - 4}}M = 9.99×10−29.99 \times {10^{ - 2}}M
pOH= -log[OH-] = -log[9.99×10−29.99 \times {10^{ - 2}}] =1.00
pH=14-1
pH=13.

The pH at the end point is 13.

Note: You must understand that titration is the method of chemical analysis where the quantity of any sample constituent is calculated by applying an exactly specified quantity of another component to the measured sample with which the target constituent responds in a definite and established proportion.