Calculate the pH

Question

Calculate the pH of a 0.35 M solution of sodium carbonate (Na2CO3). For H2CO3 Ka1 = 4.46 x 10-7 and Ka2 = 4.69 x 10-11.

Answer 1

11.94

CO

Because they gave you 2 Ka values assume this is a polyprotic acid/base This means, for example:

H₂CO₃ -----> H⁺ + HCO₃ Ka1 = 4.46 x 10^-7 *The products can be broken down further*

HCO₃ ------> H⁺ + CO₃ Ka2 = 4.69 x 10^-11

sodium carbonate is a base so you have to first determine the [OH^-] concentration from it:

Na₂CO₃ ------> 2Na⁺ + CO₃^2- *Now determine which product you can add water to to get the Ka*

water decomposition: H₂O -----> H⁺ + OH^- *H⁺ can be added to CO₃^2- to get HCO₃

H₂O + CO₃^2- -----> HCO₃^- + OH^- *now do ice table*

I 0.35 0 0

C -x +x +x

E 0.35-x x x

*remember Kb/Ka = Products/Reactants. but first we need to find Kb because we are dealing with a base*

*remember Ka x Kb = Kw Kw =10^-14

*We have Ka, which was given, we know Kw, now solve for Kb* (Use Ka2 because that's what's on above)

4.69 x 10^-11 x Kb = 10^-14 Kb = 2.13 x 10^-4

now we can use Kb = Products/Reactants

2.13 x 10^-4 = x²/0.35-x Assume x is small and don't need (but need to make sure it is valid)

2.13 x 10^-4 = x²/0.35

(2.13 x 10^-4)(0.35) = x² now square root both sides

x = .008634 *to make sure this is valid divide x by the number that was subtracted from equation above

then multiply by 100. If less than 5% then it is valid

ex. (.008634/0.35) x 100 = 2.5%

*now this x is for a base and a base has pOH*

pOH = -log(x)

pOH = -log(.008634) = 2.06

*to find the pH use the equation below and solve for pH

pH + pOH = 14

pH + 2.06 = 14

pH = 14 - 2.06

= 11.94