Part 1a buffer is prepared by mixing 462 ml of 0183 m naoh...

Part 1.A buffer is prepared by mixing 46.2 mL of 0.183 M NaOH with 131.9 mL of 0.231 M acetic acid. What is the pH of this buffer? (The pKa for acetic acid is 4.75.) Part 2.Calculate the pH of this buffer after the addition of 0.232 g of solid NaOH

Transcript

00:01 Okay, so this is a question about the buffer solution.

00:06 And let's look at it first.

00:09 You have 16 .2 milliliters of the 0 .1a3, mores of the sodium hydroxide.

00:19 From here, we can calculate the mores of the sodium hydroxide, which will be equal to the concentration times the volume.

00:28 So the concentration times the value is equal to a 40 ,000, 6 .2 times 0 .183 minus 1 ,000 which will be equal to a .45 times 10 to minus 3 more.

00:56 And then you mix that with the acidic acid, ha, right, and tells you the volume is 131 .9 mid -liter and the concentration is 0 .231 mores.

01:11 From that you can calculate the mores of the ha same cv right so would be equal to 2 31 times 1 31 .9 over 1 ,000 that would be equal to that me calculate with calculator the answer would be about 30 .47 times tend to minus three more.

01:47 So now we can write down the reaction, how they react, right? the reaction simply is sodium hydroxide, react with the ha, the acidic acid, to form the sodium acetates and h2o.

02:04 Apparently, you see the reaction ratio is one to one.

02:09 And looking at the mores here, you'll find the ha, the acidic acid is far more than the sodium.

02:17 Hydroxide so which means that all the sodium hydroxide will be consumed and to become the sodium acet so here we can know the mores of the sodium hydrostate would be equal to the mores of the sodium hydroxide and that's a 0 .45 times 10 to minus 3 and we can also calculate that the remaining acidic acid which will be equal to the initial acidic acid right which we just calculated minus the reacted one so should be equals to the most of the sodium acid tates so which would be equal to 30 .47 minus 8 .45 times 10 to minus minus 3 minus 3 here and you will get about 30 point for 7 over minus 0 .8 .45 which would be equal to 22 .02 times 10 to minus 3 and again now we know the more of the acid and the more of the base so it's a buffer solution right because there's a coexistence of the conjugate acid and the conjugate base here the ph would be equal to pk minus log, the concentration of the ha over the concentration of the acetates.

03:57 So here the ratio of the concentration should be the ratio of the morse, right? because it's the same solution.

04:05 The pk is 4 .75, minus log.

04:09 Here is 22 .02 times 10 to minus 3, over 8 .4 .5...

Question

Part 1.A buffer is prepared by mixing 46.2 mL of 0.183 M NaOH with 131.9 mL of 0.231 M acetic acid. What is the pH of this buffer? (The pKa for acetic acid is 4.75.) Part 2.Calculate the pH of this buffer after the addition of 0.232 g of solid NaOH

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