I need help on CHEMISTRY.?

Question

I need to know how to solve these problems:

1. Find the pH and pOH for a solution which contains 5.33 grams of HCL dissolved in 2.50 dm^3 of solution.

2 Find the [H+], the pH, and the percent ionization of:

a) 0.025 M solution of HCN, K= 6.2 X 10^-10
(underneath the K is an a or something.. i can't quite read it)

3. Calculate the K(a or something) for a 0.010 M solution of weak acid with pH = 5.3

4. Determine the solubility of Cu3(PO4)2 if its ksp is 1.93 X 10^-37.

The answer I got for #4 was 1.78 X 10^-8

Is that correct?

Any help would be greatly appreciated!

Answer 1

For the first one, convert the mass of HCl into moles by dividing by the molar mass of HCl. Divide that number of moles by 2.5 to get the molar concentration of HCl. Since HCl is a strong acid, the H+ concentration is equal to the concentration of HCl. Take the log of the HCl concentration and change the sign to get the pH. Calculate the pOH from the relationship:

pH + pOH = 14.

2. When you dissolve a weak acid like HCN in water, a small part of the HCN ionizes to form H+ and CN- ions. If you start with just the weak acid, then the [H+] = [CN-].

For any weak acid (assume HA) , the expression for Ka is:

Ka=[H+][A-]/[HA]

Since most weak acids are really weak, only a tiny part of the acid ionizes, so the final [HCN]=0.025.

So, Ka = 6.2 X 10^-10 = x^2/0.025

When you solve for x, that will be the [H+]. Find the pH from that.

3. Just work the problem you just did backwards. Convert the pH into [H+], and you will have:

Ka=[H+]^2/0.01

4. For this compound,

Ksp = [Cu2+]^3[PO43-]^2

If you let x equal the moles of the compound that dissolves/L of water, then the final [Cu2+]= 3x, and the [PO42-]= 2x.

So, you'll have:
1.93 X 10^-37=(3x)^3(2x)^2=108x^5
=1.78 X 10^-8

You are correct.