Ionic EquilibriumHard
Question
A volume of 20 ml of 0.8 M-HCN solution is mixed with 80 ml of 0.4 M-NaCN solution. Calculate the pH of the resulting solution. The value of Ka of HCN = 2.5 × 10−10 (log 2 = 0.3).
Options
A.9.9
B.9.3
C.4.1
D.4.7
Solution
$P^{H} = P^{K_{a}} + \log\frac{\left\lbrack CN^{-} \right\rbrack_{0}}{\lbrack HCN\rbrack_{0}}$
$= - \log\left( 2.5 \times 10^{- 10} \right) + \log\frac{80 \times 0.4/100}{20 \times 0.8/100} = 9.9$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
What will be the percentage error in measuring hydrogen ion concentration in a 10−6 M-HCl solution on neglecting the con...The self ionisation constant for pure formic acid, K = [HCOOH2+][HCOO-] has been estimated as 10-6 at room temperature. ...Consider an aqueous solution, 0.1 M each in HOCN, HCOOH, (COOH)2 and H3PO4, for HOCN, we can write Ka(HOCN) = . [H+] in ...pH at which a basic indicator with Kb = 1.0 × 10-10 changes colour when the concentration of indicator is 10-2 M :-...The dissociation constant of acetic acid is 0.000018 and that for cyano acetic acid is 0.0036 at 298 K. What would be th...