Question

calculate the concentration of $hc_{6}h_{6}o_{6}^{-}$ in an aqueous solution of $0.0686m$ ascorbic acid, $h_{2}c_{6}h_{6}o_{6}(aq)$. $hc_{6}h_{6}o_{6}^{-} = square m$. submit answer retry entire group 8 more group attempts remaining

Explanation:

Step1: Write the first dissociation equation

Ascorbic acid (\(H_{2}C_{6}H_{6}O_{6}\)) is a diprotic acid. The first dissociation is \(H_{2}C_{6}H_{6}O_{6}(aq)
ightleftharpoons H^{+}(aq)+HC_{6}H_{6}O_{6}^{-}(aq)\). Let \(x\) be the concentration of \(H^{+}\) and \(HC_{6}H_{6}O_{6}^{-}\) formed from the first dissociation. The initial concentration of \(H_{2}C_{6}H_{6}O_{6}\) is \(c = 0.0686M\). At equilibrium, \([H_{2}C_{6}H_{6}O_{6}]=(0.0686 - x)M\), \([H^{+}]=xM\), \([HC_{6}H_{6}O_{6}^{-}]=xM\). The second dissociation \(HC_{6}H_{6}O_{6}^{-}(aq)
ightleftharpoons H^{+}(aq)+C_{6}H_{6}O_{6}^{2 -}(aq)\) occurs to a much lesser extent (since \(K_{a1}\gg K_{a2}\) for diprotic acids). For ascorbic acid, \(K_{a1}=8.0\times10^{-5}\), \(K_{a2}=1.6\times 10^{-12}\).

Step2: Apply the approximation

Since \(K_{a1}\) is small, \(0.0686 - x\approx0.0686\). Using the formula for \(K_{a1}=\frac{[H^{+}][HC_{6}H_{6}O_{6}^{-}]}{[H_{2}C_{6}H_{6}O_{6}]}\), substituting the values \(K_{a1}=8.0\times 10^{-5}=\frac{x\times x}{0.0686}\).

Step3: Solve for \(x\)

$$x^{2}=8.0\times 10^{-5}\times0.0686$$

$$x^{2}=5.488\times 10^{-6}$$

$$x=\sqrt{5.488\times 10^{-6}}$$

$$x = 2.34\times10^{-3}$$

Answer:

\(2.34\times 10^{-3}\)