certain weak acid hasPka= 4find ph of 0.01m NaA(aq)

To find the pH of a 0.01M NaA(aq) solution, we first need to understand that NaA is the sodium salt of the weak acid, and it will completely dissociate in water to form Na+ and A- ions. The A- ion will react with water to form HA (the weak acid) and OH-.

The reaction can be written as follows:

A- + H2O ⇌ HA + OH-

Since we know the pKa of the weak acid, we can calculate the pKb of the base (A-) using the following relationship:

pKa + pKb = 14

So, pKb = 14 - pKa = 14 - 4 = 10

The Kb (base dissociation constant) can be calculated from pKb as follows:

Kb = 10^-pKb = 10^-10

Now, we can set up an ICE (Initial, Change, Equilibrium) table for the reaction:

Initial: [A-] = 0.01M, [HA] = 0, [OH-] = 0 Change: [A-] = -x, [HA] = +x, [OH-] = +x Equilibrium: [A-] = 0.01-x, [HA] = x, [OH-] = x

We can now write the expression for Kb:

Kb = ([HA][OH-])/[A-] = (x)(x)/(0.01-x)

Since Kb is very small, we can assume that x is much smaller than 0.01, so the equation simplifies to:

Kb = x^2/0.01

Solving for x gives the concentration of OH- ions.

x = sqrt(Kb * 0.01) = sqrt(10^-10 * 0.01) = 10^-6 M

The pOH can be calculated from [OH-] as follows:

pOH = -log[OH-] = -log(10^-6) = 6

Finally, we can find the pH using the following relationship:

pH = 14 - pOH = 14 - 6 = 8

So, the pH of the 0.01M NaA(aq) solution is 8.