What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 HCN with 50.00 mL of $0.030 \mathrm {~mol} \mathrm {~L} ^ { - 1 }$ NaCN? Assume that the volumes of the solutions are additive and that K_a = $4.9 \times 10 ^{- 10} \text { for } \mathrm { HCN } \text {. }$ A)1.5 × 10^-¹⁰ mol L^-1 B)4.9 × 10^-10 mol L^-1 C)1.6 × 10^-⁹ mol L^-1 D)7.0 × 10^-6 mol L^-1

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The correct answer is C because the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 HCN with 50.00 mL of $0.030 \mathrm {~mol} \mathrm {~L} ^ { - 1 }$ NaCN is 1.6 × 10^-⁹ mol L^-1. This can be calculated using the Henderson-Hasselbalch equation:$$pH = pK_a + \log \frac{[A^-]}{[HA]}$$where $pK_a$ is the negative logarithm of the acid dissociation constant, [A^-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.In this case, the $pK_a$ of HCN is $4.9 \times 10 ^{- 10} \text { for } \mathrm { HCN } \text {. }$, the concentration of the conjugate base (CN^-) is 0.030 mol L^-1, and the concentration of the acid (HCN) is 0.10 mol L^-1. Substituting these values into the Henderson-Hasselbalch equation, we get:$$pH = 4.9 + \log \frac{0.030}{0.10} = 4.9 - 0.52 = 4.38$$The hydronium ion concentration is then calculated using the following equation:$$[H^+] = 10^{-pH} = 10^{-4.38} = 1.6 \times 10^{-9} \text { mol L}^{-1}$$

What Is the Hydronium Ion Concentration in a Solution Prepared 0.030 mol L−10.030 \mathrm {~mol} \mathrm {~L} ^ { - 1 }0.030 mol L−1

What Is the Hydronium Ion Concentration in a Solution Prepared $0.030 \mathrm {~mol} \mathrm {~L} ^ { - 1 }$