You prepare $1.0~\mathrm{L}$ of a $0.25~\mathrm{M}$ acetic acid solution with a final $\ce{pH}$ of $6.0$. What are the molar concentrations of all relevant acetic acid species ($[\ce{HA}]$ and $[\ce{A-}]$) given that the $K_a$ for acetic acid is $1.74 \cdot 10^{-5}~\mathrm{M}$?
I am getting confused with this problem.
Since the $\ce{pH}$ is given, I know what the $[\ce{H^+}]$ is. So now when I try to do the ICE table
$$ \begin{array}{l|ccc} & \ce{HA} & \ce{H+} & \ce{A-} \\\hline \text{Initial} & 0.25 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{End} & 0.25 -x & +x & +x \\\hline \end{array}$$
And from here I begin to assume $[\ce{H^+}] = [\ce{A^-}]$, which I am not sure about. Then I set $x = 10^{-6.0} = 1\cdot 10^{-6}$ so I get $[\ce{A^-}] = 1\cdot 10^{-6}~\mathrm{M}$ and $[\ce{HA}] = 0.24999~\mathrm{M}$ which I think is incorrect, and to even further ensure my that it's incorrect, when I attempt to check the $K_a$ value with this, it does not match.
My second approach: $[\ce{HA}] = 0.25~\mathrm{M}$
In this I determine the $\mathrm{p}K_a$ from the $K_a$ which turns out to be $4.759$, which indicates that there should be more $[\ce{A-}]$ than $[\ce{HA}]$.
I now use the Henderson–Hasselbalch equation: \begin{align} 6.0 &= 4.759 + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\\ 17.40 &= \frac{[\ce{A-}]}{[\ce{HA}]}\\ 17.40 &= \frac{[\ce{A^-}]}{0.25~\mathrm{M}}\\ [\ce{A-}] &= 4.35~\mathrm{M}\\ \end{align}
I feel more confident about my second answer.
Can someone please help me out with this particular problem and perhaps tell me procedure I should use as well as what the correct answer should come out to be and why?
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