00:01
So in this question, the particular ca of benjioic acid it is given.
00:07
The ka of benjoyic acid, so that is 6 .30 into 10th power of 4.
00:13
So here, ph of a particular it is prepared by combining 50m.
00:23
Potassium benjoyate and 50m.
00:29
Of 1 molar benjoic acid.
00:34
So like this buffer it is prepared and ph of buffer we have to find out.
00:40
This is the question.
00:42
So how to find out ph of buffer? so we know that formula.
00:47
So the formula it is pca is equal to minus logarithm of ka.
00:52
That is minus logarithm of 6 .30 into 10 to the per of minus 5.
00:59
That is equal to so here the molarity molarity is equal to moles per volume so here that means for example moles moles is equal to molarity into volume molarity into volume that is volume into the liter so moles of benjoic acid how to find out moles of benjoic acid so the moles of benjoic acid that is molarity it is one molar into what is the volume here volume in liters but here 15 that is in ml 50 divided by thousand that is equal to 0 .05 molar and they in the same manner moles of potassium benzoyate moles of potassium benjureate potassium so most of potassium benjiate one molar into 50 divided by thousand molar that is very much of potassium benjade 1 molar into 50 divided by thousand molar that is very equal to 0 .05...