Question

In: Chemistry

1.2 gram NaOH is added to 1.00 liter of 0.100 M HCN. Assume no volume change....

1.2 gram NaOH is added to 1.00 liter of 0.100 M HCN. Assume no volume change.

What is the new pH?
  
HCN    + H2O        CN -       +        H3O +     K a =    6.2 x 10 -10

3.0

3.7

4.6    

8.9

7.7

10.6

9.8

5.1

Solutions

Expert Solution

Sol: HCN + NaOH ------> H2O + NaCN

No. Of moles of HCN = molarity × volume = 0.1 × 1 = 0.1 M

No. Of moles of NaOH  = mass / molar mass = 1.2/40 = 0.03 M

According to the reaction one mole of NaOH react with 1 mole of HCN.

0.03 moles of NaOH react with 0.03 moles of HCN to produce 0.03 moles of NaCN and 0.07 moles of weak acid HCN remains in the solution.

Now the solution is act as buffer in which there is a weak acid and its salt is present.

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

Pka = -log Ka = -log( 6.2 + 10^ (-10)) = 9.2

[ salt ] = no. Of moles of salt / volume = 0.03 / 1 = 0.03

[Acid ] = 0.07 / 1 = 0.07

Now, pH = 9.2 + log ( 0.03/ 0.07) = 9.2 - 0.368 = 8.83

So the correct option is 8.9


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