Answer:
A) pH of the buffer is 3.44
B) pH of the buffer solution is 3.37
Explanation:
Relation between K_{a} and pK_{a} is as follows.
[tex]pK_{a} = -log (K_{a})[/tex]
. [tex]pK_{a} = -log (K_{a}) = -log(4.5 \times 10^{-4}) = 3.347[/tex]
The relation between pH and pK_{a} is as follows.
[tex]pH = pK_{a} + log\frac{[conjugate base]}{[acid]}[/tex]
[tex]= 3.347+ log \frac{0.15}{0.12} = 3.44[/tex]
pH of the buffer is 3.44.
b)
mol of HCl added = 11.6M *0.001 L = 0.0116 mol
In the given reaction, [tex]NO^{-}_{2}[/tex] will react with [tex]H^{+}[/tex] to form[tex]HNO_{2}[/tex]
No. of moles of
[tex]NO^{-}_{2} = 0.15 M \times 1.0 L = 0.15 mol[/tex]
And, no. of moles of [tex]HNO_{2} = 0.12 M \times 1.0 L[/tex]
= 0.12 mol
after the reaction :
No. of moles of [tex]NO^{-}_{2}[/tex] = moles present initially - moles added
= (0.15 - 0.0116) mol
= 0.1384 mol
Moles of [tex]HNO_{2}[/tex] = moles present initially + moles added
= (0.12 + 0.0116)
= 0.1316 mol
As,
[tex]K_{a} = 4.5 \times 10^{-4} pK_{a} = -log (K_{a}) = -log(4.5 \times 10^{-4}) = 3.347[/tex]
Since, volume is both in numerator and denominator, we can use mol instead of concentration.
[tex]pH = pK_{a} + log \frac{[conjugate base]}{[acid]}[/tex]
= 3.347+ log {0.1384/0.1316}
= 3.369
≅ 3.37
pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37
