# PH - really basic, I know, but I'm tired

1. Apr 27, 2005

### CasanovaFrankenstein

pH -- really basic, I know, but I'm tired

Hi Chem masters out there,

Anyone want to help me with this one?

What is the pH of a solution with 10 g of sodium carbonate and 10 g of sodium bicaronate dissolved in enough water to make 0.25 mL of solution?

2. Apr 27, 2005

### Artermis

Sodium Carbonate = $$Na_2CO_3$$
Sodium Bicarbonate = $$NaHCO_3$$

0.25 mL = 0.0025L

You can ignore the Na+ ion because it's neutral, and the sodium bicarbonate releases the H proton into solution whih results in the hydronium ion being formed, therefore the concentration of the sodium bicarbonate will be equal to the concentration of the H+ ion.

Concentration of sodium bicarbonate:
10g
molar mass = 84.01 g/mol
to get moles, take grams and divide it by molar mass:
moles of sodium bicarbonate:
$$1.2 times 10^{-1}$$

divide the moles over litres of solution to get concentration:

47.6 M? okay that can't be right =/

Then you're supposed to -log(x) that to get the pH.
In case of a base, get the pOH then 14 - pOH to get pH.
I think I did something wrong. Maybe my chemical formula is wrong? I think it might form OH- actually =(

I'm sorry!

Last edited: Apr 27, 2005
3. Apr 27, 2005

### Staff: Mentor

4. Apr 27, 2005

### Artermis

*gasps*
I'm stupid =(

5. Apr 28, 2005

### GCT

just in case "HH" is the henderson hasselbach equation.

Note that 10g represents the whole ionic compound, you'll need to first find the molar mass of each compound, then convert 10g to moles. Then you'll need to use the stoichiometric ratio (which may be 1:1, you should figure it out), to find the moles of each anionic component and divide each value by the volume. Remember to use the correct units.

6. Apr 28, 2005

### CasanovaFrankenstein

Thanks Guys! That was fast and easy

7. Apr 28, 2005

### r3dxP

I like the way you solved it lol, you have to incorporate the weakacid equilibrium constant..
HCO3- => H+ + CO3^2- with Ka2 of 1.5 x 10^-11
you have (10.g)/(84.0g/mol NaHCO3) = .12mol HCO3-
and have (10.g)/(106.0g/mol Na2CO3) = .094mol CO3^2-
.12mol HCO3- / .00025L = 480M
.094mol CO3^2- / .00025L = 376M
Using Henderson-Hasselbalch Equation.. pH = pKa + log( [CO3^2-]/[HCO3-] )
you get.. pH = -log(1.5E-11) +log(376M/480M) assuming 376+x/480-x ~= 376/480.
so you will get: pH = 10.82 - .106 = 10.71 thus pH=10.71
if you dont understand something i did, tell me :p