50 g of Na2CO3 is dissolved in 4 liters of aqueous solution at 25 C. calculate the ph of the solution. Na2CO3 is a strong electrolyte, 100% ionized in aqueous solution. here's what i did:


Molarity of it will be .118

pOH wil be -log(.118) = .928

pOH + Ph = 14.00 so pH is 13.072

is this right?

You've correctly calculated the concentration of Na2CO3. Since we're assuming 100% ionization, the concentration of CO3^2- is the same. On the next step, however, you have made an error. To calculate the concentration of OH, you need to take the equilibrium reaction:

CO3^2- + H2O <-> HCO3- + OH-

into account. The Equilibrium equation governing this equation is:

Kb = [HCO3-] [OH-] / [CO3^2-] = 1.8 x 10^-4

Because there is no other source of HCO3-, we can assume that its concentration equals that of OH-. We can also assume that the the amount of CO3^2- is reduced by an amount so small by this reaction that it can be neglected. That gives us, where x is the concentration of OH-, the equation:

x^2 / 0.118 = 1.8 x 10^-4

Solve for x and use it as the concentration of OH-, to get the pOH and from it the pH as you did.