100 ml of a 0300 m solution of a weak monoprotic acid is...

Name: 100 ml of a 0300 m solution of a weak monoprotic acid is titrated with 0150 m koh after 100 ml of the koh solution is added the ph is 524 what will be the ph after another 100 ml of the koh solution i
Uploaded: 2023-07-25T01:59:39-08:00
Duration: 4 min 23 s
Channel: Angela Williamson
Description: 100 ml of a 0300 m solution of a weak monoprotic acid is titrated with 0150 m koh after 100 ml of the koh solution is added the ph is 524 what will be the ph after another 100 ml of the koh solution i

100 mL of a 0.300 M solution of a weak monoprotic acid is titrated with 0.150 M KOH. After 100 mL of the KOH solution is added, the pH is 5.24. What will be the pH after another 100 mL of the KOH solution is added? A. 8.82 B. 9.01 C. 8.62 D. 9.12 E. 10.48

Transcript

00:01 We are given 100 milliliters of 0 .3 molar unknown acid, and we add 100 milliliters of 0 .150 molar koh, and when that happens, the ph is 4 .64.

00:10 We are then asked to determine what the ph would be after another 100 milliliters of the base was added.

00:19 So we are going to first have to figure out what the ka would be for this unknown acid, and we can do that using this equation.

00:27 H plus equals ka times the concentration of the weak acid over the concentration of that base.

00:37 Make these concentration signs right here.

00:42 So we'll go ahead and plug that information in.

00:45 So two point, sorry, to find our h plus concentration, it's 10 to the negative 4 .64, and that equals 2 .29 times 10 to the negative fifth molar.

01:00 So that's where this number is coming from.

01:02 2 .29 times 10 to the negative fifth equals our ka, which is what we're trying to find, the concentration of our weak acid.

01:12 We need to put that into millimoles.

01:15 So 100 times 0 .3 is going to be 30.

01:20 So we have 30 millimoles of that in there to start with, and we have zero millimoles of that base, and then we are going to add that.

01:29 So 100 times 0 .15 is going to give us 15 millimoles, and we will subtract 15 from the top.

01:36 So 15 over 15, that's going to cancel out, and that tells us that our ka equals 2 .29 times 10 to the negative fifth...

Question

100 mL of a 0.300 M solution of a weak monoprotic acid is titrated with 0.150 M KOH. After 100 mL of the KOH solution is added, the pH is 5.24. What will be the pH after another 100 mL of the KOH solution is added? A. 8.82 B. 9.01 C. 8.62 D. 9.12 E. 10.48

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