Question

Question: Determine the pH of a 0.100 Msolution of (NH₄)₂C₂O₄.

Short answer

Answer:The pH of the solution is 6.69.

Step-by-step solution

Determination of pH of ammonium ions

The reaction for the given compound is,

$\left(N{H}_4\right)C_2{O}_4+H_{2}O\to 2N{H_4}^{+}+C_2{O}_4^{-2}+H^{+}$

The ICE table for the reaction is,

	NH4+	H+	NH3
Initial	2(0.1)M	0	0
Change	-x	+x	+x
Equilibrium	0.2-x	x	x

Substitute the above values in the equilibrium constant expression,

Determination of pOH of oxalate ions 

The reaction is,

$C_2{O}_4^{2-}+H_{2}O\to H{C}_2{O}_4^{-}+O{H}^{-}$

The ICE table for the reaction is,

	C2O4-2	HC2O4-	OH-
Initial	0.1M	0	0
Change	-x	+x	+x
Equilibrium	0.2 - x	- x	x

Substitute the above values in the equilibrium constant expression,

$$\begin{gathered} K_b=\frac{\left[H{C}_2{O}_4^{-}\right]\left[O{H}^{-}\right]}{\left[C_2{O}_4^{2-}\right]} \\ 6.46\times {10}^{-12}=\frac{x^2}{\left[0.1-x\right]} \\ x=2.54\times {10}^{-6} \end{gathered}$$

The pOH is,

$$\begin{gathered} pOH=-\log \left(2.54\times {10}^{-6}\right) \\ =5.6 \end{gathered}$$

Determination of pH

The pH of a 0.100 Msolution of (NH₄)₂C₂O₄ is,

$$\begin{gathered} 4.97+5.6=10.57 \\ pH+pOH=14 \\ pH=14-10.57=3.4 \end{gathered}$$

Hence, the pH and pOH is 6.69 and 7.31 respectively.

Thus, the pH a 0.100 Msolution of (NH₄)₂C₂O₄ is 6.69.