A 0.052M solution of benzoic acid, C6H5COOH, is titrated with a strong base. What is the [H+] of the solution half-way to the equivalence point? (Ka = 6.3x10^-5). How do you do this problem because I keep getting 7.9 x10^-3M but that’s wrong.
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Hi Manny!
This question is definitely tricky because what it’s asking is the concentration of [H+] at the half way equivalence point. When an acid is titrated to the half equivalence point, it’s H+ concentration equals the Ka. The easiest way for me to prove this is to look at the Henderson Hasselbalch equation: pH = pKa + log([A-]/[HA]). So at the half equivalence point, the concentration of the acid is equal to the concentration of the conjugate base it produced (since this is half right). Thus, let’s have x represent the concentration of the acid, pH = pKa + log([x]/[x]) so pH = pKa + log(1) and therefore, pH = pKa. We can actually take out the logarithm function from that equation and have [H+] = Ka. From here, we can solve for the concentration of [H+] and so it equals 6.3x10^-5M (the same value of the Ka)
I hope this helped and feel free to ask more questions :)
This question is definitely tricky because what it’s asking is the concentration of [H+] at the half way equivalence point. When an acid is titrated to the half equivalence point, it’s H+ concentration equals the Ka. The easiest way for me to prove this is to look at the Henderson Hasselbalch equation: pH = pKa + log([A-]/[HA]). So at the half equivalence point, the concentration of the acid is equal to the concentration of the conjugate base it produced (since this is half right). Thus, let’s have x represent the concentration of the acid, pH = pKa + log([x]/[x]) so pH = pKa + log(1) and therefore, pH = pKa. We can actually take out the logarithm function from that equation and have [H+] = Ka. From here, we can solve for the concentration of [H+] and so it equals 6.3x10^-5M (the same value of the Ka)
I hope this helped and feel free to ask more questions :)