Name: Applications of Solubility Product
Uploaded: 2019-07-19
Duration: 8 min 36 s
Description: Solubility product helps determine the solubility of sparingly soluble salts and the condition of salt precipitation. Learn more in this video.

Question

What are

[H^{+}], [A^{-}]

and

[B^{-}]

in a solution that contains

0.03 M HA

and

0.1 M HB

.

K_{a}

for

HA

and

HB

are

3.0 \times 10^{- 4}

and

1.0 \times 10^{- 10}

respectively.

Accepted Answer

When two weak acids are added in water, then, a case of simultaneous equilibria is obtained in which both the weak acids undergo dissociation in water simultaneously.

$HA ⇌ H^{+} + A^{-} K_{a 1} Concentration at t = 0 : C_{1} 0 0 Concentration at t = t_{eq} : C_{1} - C_{1} α_{1} C_{1} α_{1} + C_{2} α_{2} C_{1} α_{1} HB ⇌ H^{+} + B^{-} K_{a 2} Concentration at t = 0 C_{2} 0 0 Concentration at t = t_{eq} : C_{2} - C_{2} α_{2} C_{1} α_{1} + C_{2} α_{2} C_{2} α_{2} where C_{1} and C_{2} are initial concentrations of HA and HB and α_{1} and α_{2} are the degrees of dissociation for HA and HB respectively . K_{a 1^{}} = \frac{(C_{1} α_{1} + C_{2} α_{2}) (C_{1} α_{1})}{(C_{1} - C_{1} α_{1})} K_{a 1^{}} (C_{1} - C_{1} α_{1}) = (C_{1} α_{1} + C_{2} α_{2_{}}) (C_{1} α_{1}) - - - - (1) K_{a 2} = \frac{(C_{1} α_{1} + C_{2} α_{2}) (C_{2} α_{2})}{(C_{2} - C_{2} α_{2})} K_{a 2} (C_{2} - C_{2} α_{2}) = (C_{1} α_{1} + C_{2} α_{2}) (C_{2} α_{2}) - - - - (2) Adding (1) and (2), we get : K_{a 1} (C_{1} - {Cα}_{1} α_{1}) + K_{a 2} (C_{2} - C_{2} α_{2}) = (C_{1} α_{1} + C_{2} α_{2})^{2} K_{a 1} C_{1} (1 - α_{1}) + K_{a 2} C_{2} (1 - α_{2}) = (C_{1} α_{1} + C_{2} α_{2})^{2} Ignoring α_{1} and α_{2} with respect to 1, 1 - α_{1} = 1 and 1 - α_{2} = 1 Hence, [H^{+}] = (C_{1} α_{1} + C_{2} α_{2}) = \sqrt{(K_{a 1} C_{1} + K_{a 2} C_{2})}$

$In the given question, [HA] = 0.03 M, [HB] = 0.1 M, K_{a 1} = 3 \times 10^{- 4} and K_{a 2} = 1 \times 10^{- 10} [H^{+}] = \sqrt{0.03 (3 \times 10^{- 4}) + 0.1 (10^{- 10})} = \sqrt{9 \times 10^{- 6} + 10^{- 11}} = 3 \times 10^{- 3} (Since, 10^{- 11} < < 9 \times 10^{- 6}) Also, HA ⇌ H^{+} + A^{-} Hence, K_{a 1} = \frac{[H^{+}] [A^{-}]}{[HA]} \Rightarrow [A^{-}] = \frac{K_{a 1} [HA]}{[H^{+}]} [A^{-}] = \frac{(3 \times 10^{- 4}) C_{1} (1 - α_{1})}{3 \times 10^{- 3}} = \frac{(3 \times 10^{- 4}) 0.03}{3 \times 10^{- 3}} = 3 \times 10^{- 3} Also, HB ⇌ H^{+} + B^{-} K_{a 2} = \frac{[H^{+}] [B^{-}]}{[HB]} \Rightarrow [B^{-}] = \frac{K_{a 2} [HB]}{[H^{+}]} [A^{-}] = \frac{(10^{- 10}) C_{2} (1 - α_{2})}{3 \times 10^{- 3}} = \frac{(10^{- 10}) 0.1}{3 \times 10^{- 3}} = 3.33 \times 10^{- 9}$

What are $[H^{+}], [A^{-}]$ and $[B^{-}]$ in a solution that contains $0.03 M HA$ and $0.1 M HB$ . $K_{a}$ for $HA$ and $HB$ are $3.0 \times 10^{- 4}$ and $1.0 \times 10^{- 10}$ respectively.

Important Questions on Ionic Equilibrium

Learn and Prepare for any exam you want !

What are H+, A- and B- in a solution that contains 0.03 M HA and 0.1 M HB. Ka for HA and HB are 3.0×10-4 and 1.0×10-10 respectively.

Important Questions on Ionic Equilibrium

Learn and Prepare for any exam you want !

What are $[H^{+}], [A^{-}]$ and $[B^{-}]$ in a solution that contains $0.03 M HA$ and $0.1 M HB$ . $K_{a}$ for $HA$ and $HB$ are $3.0 \times 10^{- 4}$ and $1.0 \times 10^{- 10}$ respectively.