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A.)Calculate the pH of a buffer solution that is 0.240 M in HC2H3O2 and 0.200 M

ID: 965428 • Letter: A

Question

A.)Calculate the pH of a buffer solution that is 0.240 M in HC2H3O2 and 0.200 M in NaC2H3O2. (Ka for HC2H3O2 is 1.8×105.)

B.)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.

C.) 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 MHCl.

.D.)A 1.0-

L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is

Calculate the pH of the solution, upon addition of 0.085 mol of NaOH to the original buffer. 1.8×105

Explanation / Answer

a)

this is a buffer so:

pH = pKa + log(acetate/acid)

pH = 4.75 + log(0.2/0.24) = 4.67081

b)

this is a buffer

pH = pKa + log(M1V1/M2V2)

pH = -log(1.8*10^-4) + log((16*0.571)/(51*0.488))

pH = 3.309

c)

the equation for basic buffer

pOH = pK + log(conjugate/base)

pKb = 4.75

mmol of NH3 = MV = 0.5*1 = 0.5 mmol

mmol of NH4+ = MV = 0.2*1 = 0.2

after addition of MV = 49/1000*1 = 0.049

mmol NH3 = 0.5-0.049 = 0.451

mmol NH4+ = 0.4+0.049 = 0.449

pOH = 4.75+log(0.449/0.451) =4.748

pH = 14-4.748 = 9.252

d)

the equation

pH = pK a+ log(conjugate/acid)

pH = 4.75 + log((0.1 +0.085)/(0.1-0.085))

pH = 5.84

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