You have 700. mL of a solution that is 0.40 M HCO3- and 0.13 M CO3 -2. What mass
ID: 1026785 • Letter: Y
Question
You have 700. mL of a solution that is 0.40 M HCO3- and 0.13 M CO3 -2.
What mass of NaHCO3 or Na2CO3 is needed to change the pH of this solution to 10.50?
Please answer both questions from picture
H,CO, Kai = 4.3×10-7 You have 70o. mL of a solution that is o.40 M HCO, and o.13 M CO, a. What mass of NaHCO,(6) or Na,COs6) is needed to change the pH of this solution to 10.502 Ku = 5.6 × 10-11 b. What volume of 3.o0 M HBr (ag) or 2.00 M KOHaa) is needed to change the pH of the original solution to 10.50?Explanation / Answer
Consider the step-wise ionization of H2CO3 as below.
H2CO3 (aq) -------> H+ (aq) + HCO3- (aq); Ka1 = 4.3*10-7; pKa1 = -log Ka1 = -log (4.3*10-7) = 6.37 ……(1)
HCO3- (aq) -------> H+ (aq) + CO32- (aq); Ka2 = 5.6*10-11; pKa2 = -log Ka2 = -log (5.6*10-11) = 10.25 ……(2)
a) Determine the pH of the original buffer by using the Henderon-Hasslebach equation as
pH = pKa + log [CO32-]/[HCO3-]
= 10.25 + log (0.13 M)/(0.40 M) = 10.25 + log (0.325)
= 10.25 + (-0.488) = 9.762 9.76
Next, determine the ratio of CO32- and HCO3- in the buffer at pH 10.50. Again, use the Henderson-Hasslebach equation as
pH = pKa + log [CO32-]/[HCO3-]
=====> 10.50 = 10.25 + log [CO32-]/[HCO3-]
=====> log [CO32-]/[HCO3-] = 0.25
=====> [CO32-]/[HCO3-] = antilog (0.25) = 1.778
=====> [CO32-] = 1.778*[HCO3-] ……(3)
Note that the pH 10.50 is higher than the original pH 9.76 of the prepared buffer. We can increase the pH of the buffer (make the buffer less acidic) by increasing the amount of the conjugate base, CO32-. CO32- is obtained from Na2CO3. We however, must realize that the concentration of HCO3- in the buffer stays constant, i.e, 0.40 M. Put [HCO3-] = 0.40 M in (3) and get
[CO32-] = 1.778*[HCO3-] = 1.778*(0.40 M) = 0.7112 M.
Mole(s) of Na2CO3 present in the buffer at pH 10.50 = (700 mL)*(1 L/1000 mL)*(0.7112 M) = 0.49784 mole.
Mole(s) of Na2CO3 present in the original buffer at pH 9.76 = (700 mL)*(1 L/1000 mL)*(0.13 M) = 0.091 mole.
Mole(s) of Na2CO3 that must be added to the pH 9.76 buffer to increase the pH to 10.50 = (0.49784 – 0.091) mole = 0.40684 mole.
Molar mass of Na2CO3 = (2*22.9897 + 1*12.011 + 3*15.9994) g/mol = 105.9886 g/mol.
Mass of solid Na2CO3 that must be added to the pH 9.76 buffer to change the pH to 10.50 (assume no volume change) = (0.40684 mole)*(105.9886 g/mol) = 43.1204 g (ans).
b) Next, the change in pH is brought about by adding KOH to the original buffer (pH 9.76). KOH reacts with HCO3- as below.
OH- (aq) + HCO3- (aq) --------> CO32- (aq) + H2O (l)
As per the stoichiometric equation,
1 mole KOH = 1 mole K2CO3.
We have 0.40684 mole CO32- extra in the new buffer at pH 10.50; therefore, mole(s) of KOH added = moles of CO32- formed = 0.40684 mole.
Volume of KOH required = (0.40684 mole)/(2.00 M) = 0.20342 L = (0.20342 L)*(1000 mL/1 L) = 203.42 mL (ans).
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