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# Henderson Hasselbach to find ml of solutions watch

1. Two 0.1M phosphate solutions are used to make up a phosphate buffer with a pH of 7.45. The pKa for the sodium dihydrogenphosphate, disodium hydrogenphosphate equilibrium is 6.95. How many ml of each solution would you add together to make up a total volume of 500 ml?

What I've done so far is:

pH = pKa + log [A-]/[HA]

so

7.45=6.95 + log [A-]/[HA]

7.45-6.95 = 0.5 = log [A-]/[HA]

10^0.5 = 3.162... = [A-]/[HA].

[HA]/[A-] = 1/3.162 = 0.316

It is after this step which I become lost and I am not sure I have done it right up till now anyways. Help would be greatly appreciated.
2. (Original post by Student21)
Two 0.1M phosphate solutions are used to make up a phosphate buffer with a pH of 7.45. The pKa for the sodium dihydrogenphosphate, disodium hydrogenphosphate equilibrium is 6.95. How many ml of each solution would you add together to make up a total volume of 500 ml?

What I've done so far is:

pH = pKa + log [A-]/[HA]

so

7.45=6.95 + log [A-]/[HA]

7.45-6.95 = 0.5 = log [A-]/[HA]

10^0.5 = 3.162... = [A-]/[HA].

[HA]/[A-] = 1/3.162 = 0.316

It is after this step which I become lost and I am not sure I have done it right up till now anyways. Help would be greatly appreciated.
Both solutions have the same concentration to start with. What you have obtained here is a ratio. so out of total volume of 500mL,

volume of HA required = 1/(1+3.162) * 500 mL

volume of the conjugate base required = 3.162/(1+3.162) * 500 mL

These two volumes will then add up to 500 mL

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