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

Calculate

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

and

[{OCN}^{-}]

in a solution that contains

0.1 M HCOOH (K_{a} = 2.4 \times 10^{- 4})

and

0.1 M HOCN (K_{a} = 4 \times 10^{- 4})

.

Accepted Answer

When two weak acids are mixed to obtain an aqueous solution, it becomes a simple case of simultaneous equilibria in which both the weak acids undergo dissociation simultaneously. Here, $HCOOH$ and $HOCN$ , both are weak acids.

Let us assume that the concentrations and degree of dissociation of $HCOOH$ and $HOCN$ be $C_{1}$ and $α_{1}$ and $α_{1},$ respectively.

$HCOOH ⇌ H^{+} + {COO}^{-} t = 0 C_{1} 0 0 t = t_{eq} C_{1} - C_{1} α_{1} C_{1} α_{1} + C_{2} α_{2} C_{1} α_{1} K_{a_{1}} = \frac{(C_{1} α_{1}) (C_{1} α_{1} + C_{2} α_{2})}{(C_{1} - C_{1} α_{1})} K_{a_{1}} (C_{1} - C_{1} α_{1}) = (C_{1} α_{1} + C_{2} α_{2}) (C_{1} α_{1}) - - - - - - (1) HOCN ⇌ H^{+} + {OCN}^{-} t = 0 C_{2} 0 0 t = t_{eq} C_{2} - C_{2} α_{1} C_{1} α_{1} + C_{2} α_{2} C_{2} α_{2} K_{a_{2}} = \frac{(C_{2} α_{2}) (C_{1} α_{1} + C_{2} α_{2})}{(C_{2} - C_{2} α_{2})} K_{a_{2}} (C_{2} - C_{2} α_{2}) = (C_{1} α_{1} + C_{2} α_{2}) (C_{2} α_{2}) - - - - - - (2)$

Adding 1 and 2, we get:

$K_{a_{1}} (C_{1} - C_{1} α_{1}) + K_{a_{2}} (C_{2} - C_{2} α_{2}) = (C_{1} α_{1} + C_{2} α_{2})^{2} [H^{+}] = (C_{1} α_{1} + C_{2} α_{2}) = \sqrt{K_{a_{1}} (C_{1} - C_{1} α_{1}) + K_{a_{2}} (C_{2} - C_{2} α_{2})} = \sqrt{K_{a_{1}} C_{1} (1 - α_{1}) + K_{a_{2}} C_{2} (1 - α_{2})}$

Ignoring $α 1 and α 2$ , with respect to $1$ , we get:

$[H^{+}] = (C_{1} α_{1} + C_{2} α_{2}) = \sqrt{K_{a_{1}} C_{1} + K_{a_{2}} C_{2}}$

In the question,

$K_{a_{1}} = 2.4 \times 10^{- 4}, C_{1} = 0.1 M, K_{a_{2}} = 4 \times 10^{- 4}, C_{2} = 0.1 M [H^{+}] = \sqrt{(0.1 \times 2.4 \times 10^{- 4} + 0.1 \times 4 \times 10^{- 4})} = 0.8 \times 10^{- 2} M K_{a_{1}} = \frac{(0.8 \times 10^{- 2}) (0.1 α_{1})}{0.1 (1 - α_{1})} \Rightarrow 2.4 \times 10^{- 4} = \frac{(0.8 \times 10^{- 2}) (0.1 α_{1})}{0.1 (1 - α_{1})}$

Ignoring $α_{1}$ with respect to $1$ , we get:

$α_{1} = 3 \times 10^{- 2} [{COO}^{-}] = 0.1 α_{1} = 0.3 \times 10^{- 2} K_{a_{2}} = \frac{(0.8 \times 10^{- 2}) (0.1 α_{2})}{0.1 (1 - α_{2})} \Rightarrow 4 \times 10^{- 4} = \frac{(0.8 \times 10^{- 2}) (0.1 α_{2})}{0.1 (1 - α_{2})}$

Ignoring $α_{1}$ with respect to $1$ , we get:

$α_{2} = 5 \times 10^{- 2} [{OCN}^{-}] = 0.1 α_{2} = 0.5 \times 10^{- 2}$

Calculate $[H^{+}], [{HCOO}^{-}]$ and $[{OCN}^{-}]$ in a solution that contains $0.1 M HCOOH (K_{a} = 2.4 \times 10^{- 4})$ and $0.1 M HOCN (K_{a} = 4 \times 10^{- 4})$ .

Important Questions on Equilibrium

Learn and Prepare for any exam you want !

Calculate H+, HCOO- and OCN- in a solution that contains 0.1M HCOOH Ka=2.4×10-4 and 0.1M HOCN Ka=4×10-4.

Important Questions on Equilibrium

Learn and Prepare for any exam you want !

Calculate $[H^{+}], [{HCOO}^{-}]$ and $[{OCN}^{-}]$ in a solution that contains $0.1 M HCOOH (K_{a} = 2.4 \times 10^{- 4})$ and $0.1 M HOCN (K_{a} = 4 \times 10^{- 4})$ .