00:01
Question is calculate the ph of a 0 .05 molar solution of benzoic acid and ca is given and also calculate percentage of the acid that has ionized in the solution.
00:15
Okay.
00:17
So in the question we have initial concentration that is equal to 0 .05 and ka is equal to 1 .2 into 10 rate to the power minus.
00:31
Okay, and benzoic acid will be dissociate in the water like this.
00:42
Equilibrium will establish and c6h5 c .o .o.
00:48
Minus plus h3 plus.
00:53
First we have to calculate the concentration of acid, salt and hydrogen ion with the help of ice table.
01:02
So we will draw ice table.
01:05
Here will come acid and here will come salt and here will come hydrogen ion.
01:18
Correct.
01:19
At t equal to zero initially.
01:22
Initially acid is present and the concentration of this is 0 .05 molar.
01:29
Salt concentration is zero, hydrogen ion concentration 0.
01:33
Okay.
01:35
So what is the change? change is minus x x x because concentration of reactant is decreasing as the reaction forward goes and concentration at equilibrium at equilibrium that will be 0 .05 minus x x and x correct so what is the expression of ionization constant it will be k equal to ratio of concentration of salt and hydrogen ion to the concentration of react that is acid so in the question we have value of k a that is equal to 1 .2 into 10 ratio of power minus 4 x into x divide by 0 .05 minus x okay okay here, value of x is very, very less than, is very, very less than 0 .05 because benzoic acid is a very weak acid, so it will ionize very, very less...