Chemistry 12th Edition

Published by McGraw-Hill Education

Chapter 15 - Acids and Bases - Questions & Problems - Page 711: 15.46

Answer

$[HA]_{initial} = 2.26 \times 10^{-3}M$

Work Step by Step

1. Find $[H^+]$ $[H^+] = 10^{-pH} = 10^{-3.26} = 5.5 \times 10^{-4}M$ 2. Using the ICE table for the first interaction of the acid with water: $[HA] = [HA]_{initial} - x$ $[H^+] = [A^-] = x$ - We know that $[H^+] = 5.5 \times 10^{-4}$, so: $x = 5.5 \times 10^{-4}$ 3. Use the Ka equation to find the $[HA]$ in equilibrium: Ka formic acid$: 1.77 \times 10^{-4}$ $1.77 \times 10^{-4} = \frac{[H^+][A^-]}{[HA]} = \frac{(5.5 \times 10^{-4})^2}{[HA]}$ $[HA] = \frac{3.025 \times 10^{-7}}{1.77 \times 10^{-4}}$ $[HA] = 1.71 \times 10^{-3}$ 4. Find the $[HA]_{initial}$ $[HA] = [HA]_{initial} - x$ $1.71 \times 10^{-3} = [HA]_{initial} - 5.5 \times 10^{-4}$ $[HA]_{initial} = 2.26 \times 10^{-3}M$

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