1)
Calculate the pH of the solution that results from each of the following mixtures.
A)
150.0 mL of 0.26 M HF with 225.0 mL of 0.32 M NaF
B)
165.0 mL of 0.12 M C2H5NH2 with 275.0 mL of 0.22 M C2H5NH3Cl
2)
A)
As a technician in a large pharmaceutical research firm, you need to produce 100. mLof a potassium dihydrogen phosphate buffer solution of pH = 6.91. The pKa of H2PO4− is 7.21.
You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O.
How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)
B)
If the normal physiological concentration of HCO3− is 24 mM, what is the pH of blood if PCO2 drops to 31.0 mmHg ?
3)Which pair of concentrations results in the most effective buffer?
0.95 MHA; 0.05 M A−
0.10 MHA; 0.90 M A−
0.20 MHA; 0.20 M A−
0.70 MHA; 0.70 M A−
1) A) Moles of HF, a weak acid = 0.15 * 0.26 = 0.039 and moles
of F-, its conjugate base = 0.225 * 0.32 = 0.072
Total volume of the mixture = 0.15 + 0.225 L = 0.375 L
Hence, New [HF] = 0.039/0.375 = 0.104 M and New [F-] =
0.072/0.375 = 0.192 M
pH of a weak acid buffer = pKa + log([HF]/[F-])
= -log(6.6*10-4) + log[(0.104)/(0.192)]
[Ka value of HF = 6.6 * 10-4]
= 2.91
B) Moles of C2H5NH2 = 0.165 *
0.12 = 0.019 and moles of
C2H5NH3+ = 0.275 * 0.22
= 0.06
Total volume = 0.44 L
Thus, [C2H5NH2] = 0.019/0.44 =
0.043M and [C2H5NH3+] =
0.06/0.44 = 0.136M
pOH of a weak base buffer = pKb +
log([C2H5NH3+]/[C2H5NH2]
)
= -log (4.3*10-4) + log[(0.136)/(0.043)]
= 3.86.
Thus, pH = 14-3.86 = 10.14
2.A) pH of the buffer = pKa +
log[HPO4-2]/log[H2PO4-]
i.e. 6.91 = 7.21 +
log[HPO4-2]/[H2PO4-]
;
[HPO4-2]/[H2PO4-]
= 0.5
Let the volume of KH2PO4 be x L. Thus, volume
of K2HPO4= 0.25-x L.
Now,
moles of KH2PO4 = 2 and moles of
K2HPO4 = 1.5
Thus,
[HPO4-2]/[H2PO4-]
= 0.5 = (1.5/0.25-x) / (2/x)
Solving for x, we obtain
x, i.e. volume of KH2PO4 = 0.1 L = 100
mL.
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