Calculate the pH of a buffet solution prepared by dissolving 4.2g of NaHC03 and

Question

Calculate the pH of a buffet solution prepared by dissolving 4.2g of NaHC03 and 5.5g of Na2C03 in 0.20 L of water. Express your answer using two decimal places. Will the pH change if the solution volume is increased by a factor of 10? A certain weak acid, HA, with a Ka value of 5.61 Times 10-6, is titrated with NaOH A solution is made by mixing 7.00mmol (millimoles) of HA and 3.00mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places. More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 73.0mL ? Express the pH numerically to two decimal places.

Explanation / Answer

Item 8:

Part A:

Calculate moles of NaHCO3 and moles of Na2CO3 used. Divide each by 0.200 L to calcualte the molar concentration of each.

Then, look up pKa for HCO3- <---> H+ + CO32-

pH = pKa + log [CO32-]/[HCO3-]

OR, look up Ka for ionization of bicarbonate. Then,

Ka = [H+][CO32-]/[HCO3-]

Plug in values and calculate [H+]

From [H+], calculate pH = - log [H+]

PART B:

yes, pH will decrease

Item 11

Part A:

pKa = 5.25 + log 3.00/ 7.00 - 3.00=4.88

moles A- = 7.00 x10^-3
[A-]= 7.00 x 10^-3 / 0.080 L=0.0875 M

Kb = Kw/Ka = 1.78 x 10^-9 = x^2 / 0.0875-x
x = [OH-]= 1.25 x 10^-5 M
pOH = 4.90
pH = 9.10

Part B:

No change...but not sure of this

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