Calculate the ph at the following points in the titration of...

Name: calculate the ph at the following points in the titration of 3000 ml of 0150 m c6h5cooh with 0150 m naoh the ka of c6h5cooh is 63x10 5 a before the addition of any naoh b after the addition of 1000 ml
Uploaded: 2023-06-08T00:36:39-08:00
Duration: 2 min 59 s
Channel: Supreeta N
Description: calculate the ph at the following points in the titration of 3000 ml of 0150 m c6h5cooh with 0150 m naoh the ka of c6h5cooh is 63x10 5 a before the addition of any naoh b after the addition of 1000 ml

Calculate the pH at the following points in the titration of 30.00 ml of 0.150 M C6H5COOH with 0.150 M NaOH. The Ka of C6H5COOH is 6.3x10-5 a. Before the addition of any NaOH b. After the addition of 10.00 ml of 0.150 M NaOH c. After the addition of 15.00 ml of 0.150 M NaOH d. After the addition of 30.00 ml of 0.150 M NaOH e. After the addition of 35.00 ml of 0.150 M NaOH f. Sketch the titration curve

Transcript

00:01 Hi, in this question we are given that benzoic acid and sodium hydroxide are titrated and they produce sodium benzoate and water.

00:07 We are given with concentration and volume of benzoic acid and sodium hydroxide in this manner and we are asked to calculate the ph of solution.

00:15 First of all, we will calculate the moles.

00:18 It is equal to molarity into volume.

00:20 We can substitute the values for both here in this manner.

00:24 Here we can eliminate the liter unit.

00:26 On solving we are getting moles of benzoic acid 0 .00450 mole and moles of sodium hydroxide 0 .00525 mole.

00:36 Here we can see 1 mole of benzoic acid reacts with 1 mole of sodium hydroxide.

00:42 Therefore, 0 .00450 moles of benzoic acid will react with 0 .00450 moles of sodium hydroxide.

00:51 In that case, we will be remaining with the sodium hydroxide.

00:56 Hence, we can calculate the remaining moles of naoh.

01:04 It will be equal to 0 .00525 -0 .00450 moles.

01:15 On solving this we are getting the remaining moles of naoh that are present in excess amount equals to 0 .00075 mole.

01:24 Now we can calculate the total volume of solution that is equal to volume of benzoic acid plus volume of sodium hydroxide.

01:32 On adding this we are getting the total volume as 0 .065 liter.

01:37 We can calculate the concentration of sodium hydroxide as we can say that the concentration of naoh will be equal to moles over total volume...

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