Multiple Choice
If the theoretical yield of a reaction is 80.0 g and the percent yield is 95%, what is the actual yield?
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3. Chemical Reactions
Percent Yield
Multiple Choice
What is the percent ionization of a 0.300 M solution of formic acid (HCOOH), given that Ka = 1.78 × 10^{-4}?
A
0.24%
B
8.7%
C
24%
D
2.4%
Verified step by step guidance1
Write the ionization equilibrium expression for formic acid (HCOOH): it partially ionizes in water as HCOOH \( \rightleftharpoons \) H\(^+\) + HCOO\(^-\).
Set up an ICE table (Initial, Change, Equilibrium) for the concentrations: Initial concentration of HCOOH is 0.300 M, and initial concentrations of H\(^+\) and HCOO\(^-\) are 0.
Define the change in concentration of HCOOH that ionizes as \( x \). At equilibrium, the concentration of HCOOH is \( 0.300 - x \), and the concentrations of H\(^+\) and HCOO\(^-\) are both \( x \).
Write the expression for the acid dissociation constant \( K_a \):
\( K_a = \frac{{[\text{H}^+][\text{HCOO}^-]}}{{[\text{HCOOH}]}} = \frac{{x^2}}{{0.300 - x}} \)
Assuming \( x \) is small compared to 0.300, approximate the denominator as 0.300 and solve for \( x \). Then calculate the percent ionization using the formula:
\( \text{Percent ionization} = \frac{x}{0.300} \times 100\% \)
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