Formic acid, HFor, has a Ka value of 1.8 x 10-4. You need to...

Formic acid, HFor, has a Ka value of 1.8 x 10-4. You need to prepare 100 mL of a buffer having a pH of 3.25 from 0.10 M HFor solution and a 0.10 M NaFor solution. How many mL of NaFor and HFor should be mixed to make the desired buffer? mL NaFor mL HFor When five drops of 0.10 M NaOH were added to 20 mL of the buffer in question 4, the pH

Transcript

00:01 In this question, they have given the formic acid.

00:05 Formic acid.

00:07 So the formic acid, it having the ca value, that is 1 .8 into 10 to the power of minus 4.

00:14 Here we have to prepare 100 ml of a buffer.

00:19 So buffer, it having the ph 3 .25 from 0 .1 molar, h, that is formic acid.

00:34 So that is formic acid.

00:36 So from 0 .1 molar formic acid, you have to prepare 100m -buffer which having the ph 3 .25.

00:46 And along with that of formic acid solution, we have to use 0 .10 sodium format.

00:54 0 .10m sodium format.

00:59 So by using this, we have to prepare the sodium format solution.

01:07 So here how many ml of sodium format, sodium format as well as formic acid, sodium format as well as formic acid? so they should be mixed to get the desired buffer.

01:27 That is the question.

01:29 Here we have to find out the how many ml or desired buffer.

01:34 That is the question.

01:35 We'll see the answer here.

01:37 So now here the molarity, molarity.

01:45 So we know that molarity is the formula of molarity that is equal to number of moles by volume in liters.

01:53 And the ca of formic acid, it is given 1 .8 into 10 to the power of minus 4.

02:02 The value it is given.

02:04 And the polarity of formic acid, molarity of formic acid, molarity of formic acid, it is given the 0 .1 molar and the volume of formic acid it is given volume of formic acid we have to find out let it be v1 and moles of formic acid so what will be the moles of formic acid so what will be the moles of formic acid it becomes number of moles into volume number of moles it is 0 .1 and v1 so this is the molcer so this is the molcer and the molarity of sodium format that is h -c -o -n -a that it is 0 .1 molar volume of sodium format that thing we have to find out volume of sodium format that is h -c -o -n -a we have to find out let it be v2 now moles of sodium format that it will be 0 .1 v2.

03:15 Now, volume of buffer solution.

03:19 So what is the volume of buffer solution we have to prepare that is equal to 100m? so, that means v1 plus v2 is equal to 100m.

03:30 Now, molarity of, molarity of h, that is formic acid in buffer.

03:40 So that it will be 0 .000.

03:45 0 .1 v1 by 0 .1...

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