Question

In: Chemistry

A. Calculate the pH of a solution formed by combining 0.125 mL of 0.55 M HI...

A. Calculate the pH of a solution formed by combining 0.125 mL of 0.55 M HI with 0.400 mL of 0.14 M KOH at 25°C.
B. Calculate the pH of a solution formed by combining 0.125 mL of 0.55 M HI with 0.400 mL of 0.18 M KOH at 25°C.
C. A 50.00-mL sample of 0.0750 M HCN was titrated with 0.3995 M NaOH. What is the pH of the resulting solution after 5.00 mL of NaOH were added at 25°C? Ka for HCN is 6.2x10–10 at 25°C.
D. A 50.00-mL sample of 0.225 M HCO2H was titrated with 0.1125 M NaOH. What is the pH at the equivalence point at 25°C?  Ka for HCO2H is 1.8x10–4 at 25°C.
E. A 25.00-mL sample of 0.225 M HNO2 was titrated with 0.1111 M NaOH. What is the pH at the equivalence point at 25°C? Ka for HNO2 is 7.2x10–4 at 25°C.
F. A solution of 0.600 g of diprotic acid in 100.0 mL of water was titrated with 0.300 M NaOH to the second equivalence point. The volume of the base used was 34.5 mL. What is the molar mass of acid in g/mol?
G. The solubility of Gd2(SO4)3 is 4.0 g/100 mL. What is the Ksp?
H. What is the molarity of Ba2+ in a saturated solution of Ba3(AsO4)2? (Ksp = 1.1 x 10-13)
I. What is the concentration of Mg2+ when MgF2 (Ksp = 6.4 x 10-9) begins to precipitate from a solution that is 0.20 M F-?
J. What is the pH of a saturated solution of Pb(OH)2 (Ksp = 1.2 x 10-5)?

please answer showing steps ( A to J )

Solutions

Expert Solution

A)

millimoles of HI = 0.125 x 0.55 = 0.06875

millimoles of KOH = 0.400 x 0.14 = 0.056

[H+] = 0.06875 - 0.056 / (0.125 + 0.400) = 0.0243 M

pH = -log (0.0243)

pH = 1.62

B)

[OH-] = 0.400 x 0.18 - (0.125 x 0.55) / 0.125 + 0.400

           = 0.00619 M

pOH = -log (0.00619) = 2.208

pH = 11.79

C.

mmoles of HCN = 50 x 0.0750 = 3.75

mmoles of NaOH = 5 x 0.3995 = 1.9975

HCN    +   NaOH    ---------------> CN-    +   H2O

3.75          1.9975                            0             0

1.75              0                                1.9975

pH = pKa + log [salt / acid]

    = 9.21 + log [1.9975 / 1.75]

pH = 9.26


D.

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (3.74 + log 0.075)

pH = 8.31


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