# A 4.7 g sample of NaHCO3 is dissolved in 150 mL of solution. What is the pH of the solution? pKa1 = 6.37 and pKa2 = 10.25.

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When a salt of acid and base is dissolved in a solution (of water), the conjugate acid or base will hydrolyze. To further explain that, let us use the salt given in the problem. When NaHCO3 is dissolved in water, it splits into Na+ and HCO3- ions.

NaHCO3 --> Na+ + HCO3-

Bicarbonate (HCO3-) ion will hydrolyze to form the acid.

HCO3- + H2O <--> H2CO3 + OH-

Looking at the equation 2, we can see now that it is a basic solution.

in this case we will now use the kb of the reaction. Since we do not have the value of kb in the problem, we can derive it using the expression:

kw = ka*kb

kb = kw/ka

kb = 10^-14/10^-6.37

kb = 10^-7.63 = 2.322x10^-8

Now, we have to get the concentration of the salt solution in order to solve the problem. Remember that:

concentration NaHCO3 = concentration of HCO3-

or, [NaHCO3] = [HCO3-]

[NaHCO3] = (4.7g/84 g-mol-)/(150/1000)L

[NaHCO3] = 0.373 = [HCO3-]

Next we have to write the equilibrium expression

kb = [H2CO3][OH-]/[HCO3-]

kb = [x][x]/[HCO3-]-x

10^-7.63 = x^2 /0.373-x

To simplify the expression, we can let x = 0 in the denominator.

10^-7.63 = x^2 /0.373

x^2 = (10^-7.63) x (0.373)

x = sqrt ((10^-7.63) x (0.373))

x = 9.351x10^-5 = [OH-]

pOH = -log [OH-]

pOH = -log [9.351x10^-5]

pOH = 4.029

pH + pOH = 14

pH = 14 - 4.029

**pH = 9.97 the pH of the solution**