What is the ph of a buffer prepared by adding 0708 mol of...

What is the pH of a buffer prepared by adding 0.708 mol of the weak acid HA to 0.609 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10^(-7). Express the pH numerically to three decimal places. pH = Part B: What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. Express the pH numerically to three decimal places. pH = Part C: What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base. Express the pH numerically to three decimal places. pH =

Transcript

00:01 Now let's look at the ph of a buffer prepared by adding 0 .708 moles of a weak acid to 0 .60 moles of a sodium of a salt of the weak acid in a two liter solution.

00:15 So first, that's the first part of the question.

00:18 Let's say the weak acid, the concentration will be 0 .708 divided by the volume 2 .00 and you have 0 .354 molar while the concentration of the conjugate base would be 0 .609 divided by 2 .00 and that will give you 0 .305 molar.

00:42 So the ph, right, the ph is basically the pka of the acid plus the log of the concentration of the base divided by the concentration of the acid itself.

00:53 The pka of the acid is minus log of its ka, that's 5 .6 exponential, minus 7 plus the log of 0 .305 divided by 0 .354 and of course this will give you 6 .185.

01:15 We're actually expressing three decimal places.

01:19 The b part, now if you were to add hcl to that solution, right, first the number of the concentration of the hcl that you added, the concentration of the hcl or the acid, right, is going to be its moles, the number of moles that you added, 0 .15 divided by 2.

01:46 That's the volume of the 0 .075 molar.

01:50 Now what happens is that when you add an acid to a buffer, the conjugate base of the buffer would react with the strong acid.

01:59 So it would mean that the concentration of a minus is going to reduce.

02:04 It will become 0 .305 minus 0 .075 and that will give you 0 .230.

02:18 And what the concentration of the weak acid will increase.

02:23 So 0 .354 plus 0 .075 and this becomes 0 .429...