00:01
In this question, we're asked to calculate the ph of the solution's given.
00:06
For the first one, calcium hydroxide will dissolve in solution.
00:12
And you see that each mole of calcium hydroxide gives us two moles of hydroxide.
00:17
So we can find the molarity of the hydroxide ion.
00:20
We can find p .o .h using negative log hydroxide ion concentration.
00:25
Ph would just be 14 minus the p .o .h.
00:28
So then for b, we have chloroacetic acid and we're given the ka.
00:35
So we're going to use an ice table.
00:37
The initial concentrations of the hydrogenium ion and the conjugate base are zero.
00:42
Then to reach equilibrium, the acid goes down by x and then the hydronium ion and conjugate base go off by x.
00:50
Then we are going to write the expression for ka.
00:56
So ka is just going to be the hydrogen ion concentration.
01:05
Times the conjugate base concentration over the concentration of the acid.
01:18
So then we fill in the values for the equilibrium concentrations.
01:25
And for this one, we can't assume x is small because ka is greater than 10 to the negative four...