00:01
Okay, so we have to calculate the h3o plus and ph of each of the polyprotech acidic solution.
00:09
For part a, we have mh3, p -o -4 with a molarity of 0 .35.
00:16
So let's create our ice table.
00:17
We have the concentration of h3, p -o -4, as well as the concentration of h2, p -o -4 minus, and the concentration of h3o -plus, and the label i -c -c.
00:32
In our initial value is 0 .350, 0 ,0, our reactants have a negative change, and our products have a positive change.
00:40
Summing b, so we get the following.
00:45
So our k value is equal to, by definition, the product concentration of our product, so h2p04 minus times h3o plus, divided by h3p04, which is equal to x times x divided by 0 .35 minus x.
01:10
So x squared divided by 0 .350 minus x, which is equal to 7 .5 times 10 to negative 3.
01:19
Okay, now let's solve for our x value.
01:22
We're going to multiply both sides by 0 .35 minus x.
01:25
So we get x squared is equal to 7 .5 times 10 to negative 3 times 0 .35 minus x, multiplying that out and then moving everything to our left...