Consider how best to prepare one liter of a buffer solution with pH = 11.22 usin

Question

Consider how best to prepare one liter of a buffer solution with pH = 11.22 using one of the weak acid/conjugate base systems shown here. Weak Acid Conjugate Base Ka pKa HC2O4- C2O42- 6.4 x 10-5 4.19 H2PO4- HPO42- 6.2 x 10-8 7.21 HCO3- CO32- 4.8 x 10-11 10.32 How many grams of the potassium salt of the weak acid must be combined with how many grams of the potassium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams potassium salt of weak acid =

grams potassium salt of conjugate base=

Explanation / Answer

here the suitable buffer is HCO3- and CO32- .

pKa = 10.32

pH = 11.22

total moles of buffer = 1 x 1 = 1.00

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

11.22 = 10.32 + log [CO32- / HCO3-]

[CO32- / HCO3-] = 7.943

[CO32- + HCO3-] = 1.00

7.943 HCO3- + HCO3- = 1

moles of HCO3- = 0.1118

moles of CO32- = 0.8882

moles of KHCO3 = 0.1118 mol

moles of K2CO3 = 0.8882 mol

mass of potassium salt of weak acid = 11.2 g

mass of potassium salt of conjugate base= 123 g

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