What mass of sodium acetate nach3co2 must be added to 0750 l...

What mass of sodium acetate (NaCH3CO2) must be added to 0.750 L of 0.134 M solution of acetic acid to make a buffer of pH = 5.00? The Ka of acetic acid is 1.8E-5. ___________g How many moles of acid are required to change the pH of the solution in part a by 0.05 pH units? ___________mol Suppose 5.0 mL of 12.0 M HCl solution is added to the buffer solution in part a. Calculate the new pH. ___________.

Transcript

00:01 To determine the mass we can first determine the moles using the henderson -hasselbalch equation, where ph is equal to pka of acetic acid, in this case, plus the log of...

00:12 A lot of people will do molarity base over molarity acid, but it's easier to just do moles.

00:19 So it'll be moles of the acetate, which is the base, over the moles of the acid, or acetic acid.

00:30 So we want to achieve a ph of 5 .00.

00:33 We'll set that equal to pka, which is the negative log of the ka value for acetic acid, the negative log of 1 .8 times 10 to the negative 5, and then plus the log of the moles of acetate that we need to add, which is what we want to calculate, divided by the moles of acetic acid that we start with.

01:01 We started with 0 .750 liters at a concentration of 0 .134 moles per liter.

01:13 Now we solve for x, and we get 1 .0 .1005 when we calculate the negative log and subtract it from both sides, is equal to the log of x divided by...

01:34 Whoops, got my numbers mixed up.

01:38 This is actually 0 .2553, and that's equal to the log of x divided by this times this gives us 0 .1005.

01:52 Then we take the antilog base 10 on both sides, that's 10 to the 10 to the 0 .2553, gives us 1 .80, and that'll be equal to x divided by 0 .1005.

02:09 So x is going to be equal to 0 .1809, and that's moles of acetic acid.

02:21 We can then calculate the grams of acetic acid by using the molar mass, sorry not grams, acetic acid, grams sodium acetate by using the molar mass of sodium acetate, which is 1 mole nac2h3o2, which gives us 1 mole acetate is 82 .034 grams, and we get then as a mass that we need to add 14 .8 grams sodium acetate.

03:13 Then when we add a strong acid such as hydrochloric acid, the hydronium ion is going to react with the acetate to produce more acetic acid and water.

03:31 So if we want to change the ph by 0 .05 units, when we add an acid it's going to decrease by 0 .05 ph units, so it's going to go to 9, oops, 4 .95, and that'll be equal to our pka, negative log of 1 .8 times 10 to the negative 5 plus the log of the moles of acetate that we have before adding the strong acid is going to be what we calculated here, 0 .1809, and then we're going to add the acid...

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