A.)Calculate the pH of the buffer that results from mixing 51.0 mL of a 0.488 M

Question

A.)Calculate the pH of the buffer that results from mixing 51.0 mL of a 0.488 M solution of HCHO2 and 16.0 mL of a 0.571 M solution of NaCHO2. The Kavalue for HCHO2 is 1.8×104.

B.)Consider a buffer solution that is 0.50

M in NH3and 0.20 M in NH4Cl. For ammonia, pKb=4.75.

Calculate the pH of 1.0 L of the solution, upon addition of 49.00 mL of 1.0

C.)A 36.0

mL sample of 0.162 M HNO2 is titrated with 0.226 M KOH. (Ka for HNO2 is 4.57×104.)

Determine the pH at the equivalence point for the titration of HNO2 and KOH.

D.) A 1.0-

L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for

Calculate the pH of the solution upon addition of 19.1 mL of 1.00 MHCl to the original buffer.

HC2H3O2 is
1.8×105.MHCl

Explanation / Answer

A )

Ka = 1.8 x 10^-4

pKa = -logKa = -log (1.8 x10^-4) = 3.74

pH = pKa + log [NaCHO2 / HCHO2]

pH = 3.74 + log (16 x 0.571 / 51 x 0.488)

pH = 3.30

part B)

Calculate the pH of 1.0 L of the solution, upon addition of 49.00 mL of 1.0 ??????

acid or base you did not give .

part C)

at equivalence point : millimoles of acid = millimoles of base

36 x 0.162 = 0.226 x V

5.832 =  0.226 x V

V = 25.8 mL

at equivalence point only salt is formed   

salt concentration = C = 5.832 / (25.8 + 36) = 0.09437 M

pKa = 3.34

pH = 7 + 1/2 [pKa + logC]

pH = 7 + 1/2 [3.34 + log 0.09437]

pH = 8.16

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