Answer:
1.62 g
Explanation:
Given that:
Concentration of HCN = 0.119 M
Assuming the ka 4.00 × 10⁻¹⁰
The pKa of HCN (hydrocyanic acid) = -log (Ka)
= - log ( 4.00 × 10⁻¹⁰)
= 9.398
pH of buffer = 8.809
Using Henderson Hasselbach equation:
[tex]pH = pKa + log \dfrac{[conjugate\ base ]}{acid}[/tex]
[tex]pH = pKa + log \dfrac{[CN^-]}{[HCN]}[/tex]
[tex]8.809 = 9.398 +log \dfrac{[CN^-]}{[HCN]}[/tex]
[tex]log \dfrac{[CN^-]}{[HCN]}= 8.809 - 9.398[/tex]
[tex]log \dfrac{[CN^-]}{[HCN]}= -0.589[/tex]
[tex]\dfrac{[CN^-]}{[HCN]}= 0.2576[/tex]
[CN^-] = 0.2576[HCN]
[CN^-] = 0.2756 (0.119) L
[CN^-] = 0.033 M
∴
The amount of NaCN (sodium cyanide) is calculated as follows:
[tex]= 1.00 L \times \dfrac{0.033 \ mol \ NacN }{1 \ L } \times \dfrac{49.01 \ g}{1 \ mol \ of \ NacN}[/tex]
= 1.62 g
