M = mols/L. You want 0.073M HClO and you want 0.250 L so that is 0.073 = mols/0.250 so mols = approx 0.01825 (I am using more significant figures than allowd---a few guard digits). Basically you want 18.25 mmols HClO in the solution.
For the base, you want mmols 250 x 0.046 = about 11.5 mmols NaClO.
We want to know how much HClO to start with.
.........HClO + NaOH ==> NaClO + H2O
I.........X......0.........0
add..............y.........y
C........-y......-y........y
E.......X-y......0.........y
We know y = 11.5 mmols NaOH to produce 11.5 mmols NaClO. If X-y = 18.2 then X = 18.25+y or 18.25+11.5 = 29.75 mmols HClO we must start with.
Since M = mmols/mL and we want 29.75 mmols, we must mL of the 0.365 stuff = 29.75/0.365 = about 81.5 mL of the 0.365 M stuff. You can check that out this way.
You will take 81.5 mL of 0.365M HClO and add 11.5 mL of the 1M NaOH. That will produce 11.5 mmols NaClO and leave (81.5 x 0.365 = 29.75)-11.5 = 18.25 mmols HClO
(HClO) = 18.25/250 = 0.073M which is what you wanted.
(NaClO) = 11.5/250 = 0.046 which is what you wanted.
A student was required to prepare 250.0 mL of a hypochlorous acid/sodium hypochlorite buffer in which the concentration of the weak acid component was 0.073 M and the concentration of the conjugate base was 0.046 M. The student was supplied with 0.365 M hypochlorous acid and 1.0M NaOH to perform this task. What volume (in L) of the acid would the student need to prepare this buffer solution?
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