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A weak acid HA after treatment with 12 mL of 0.1 M strong base BOH has a pH of 5 . At the end point , the volume of same base required is 26.6 mL. The Ka of acid is ? ​ a. 8.219 × 10 -5 ​ b. 8.219 × 10 -7 c. 8.219 × 10 -6 d. 8.219 ×10 -4 Ans ;- ( C ) 8.219 × 10 -6 Please explain in detail......

A weak acid HA after treatment with 12 mL of 0.1 M strong base BOH has a pH of 5. At the end pointthe volume of same base required is 26.6 mL. The Ka of acid is ?
​ a.  8.219 × 10-5
​ b.  8.219 × 10-7
c.  8.219 × 10-6
d.  8.219 ×10 -4
Ans ;- (C)  8.219 × 10-6
Please explain in detail......

Grade:12th pass

1 Answers

Arun
25750 Points
4 years ago
Dear student
 

As learnt in

Buffer Solutions -

The solutions which resist change in pH on dilution or with addition of small amounts of acid or alkali are called buffer solution.  

- wherein

Many body fluids  e.g.  blood

 

 Total milliequivalents of acid = milliequivalents of base

                = 26.6 x 0.1 = 2.66

But the resultant mixture has HA left

The reaction is as follows:

            HA    +    BOH \rightarrow BA    +    H2O

t=0      2.66            12         0              0

t1        1.46             0         1.2           1.2

Therefore for a buffer solution,

pH = pKa+\log \frac{\left [ salt \right ]}{\left [ acid \right ]}

\Rightarrow 5 = pKa+\log \frac{1.2}{1.46}

\Rightarrow \log Ka = -5.86

\Rightarrow Ka = 8.21 \times 10^{-6}


 

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