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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