Which of the following is a buffer solution?
A. HCl and NaCl
B. HF and NaOH
C. HCl and NaOH
D. HCN and NaCN
E. NH3 and HF
D. HCN and NaCN form a buffer solution, because it contains a weak acid and its conjugate base.
A. isn't a buffer solution because HCl is a strong acid.
B. isn't a buffer solution because NaOH is a strong base.
C. isn't a buffer solution because it contains both a strong acid and strong base.
E. isn't a buffer solution because though it contains weak acid and a weak base, they aren't conjugates of each other.
Calculate the pH of the solution after the following are mixed: 300.0 ml of 0.200 M KOH and 200.0 ml of 0.400 M HC2H3O2 (Ka = 1.8 x 10-5).
R HC2H3O2 + KOH ---> H2O + KC2H3O2
I 80.0 mmol 60.0 mmol lots 0
C -60.0 -60.0 +60.0 +60.0
E 20.0 ~0 lots 60.0
Weak Acid-base conjugate pair present
pH = pKa + log(conj base/acid)
pH = -log(1.8x10-5) + log(60.0/20.0)
pH = 5.22
What is the pH of a solution consisting of 0.75 M HC2H3O2 and 0.50 M NaC2H3O2? Ka of HC2H3O2 is 1.8 x 10-5.
This is a buffer solution, because we have a significant amount of both a weak acid and its conjugate base. Therefore, we can use the Henderson-Hasselbalch equation.
pH = pKa + log (Ac-/HAc)
pH= 4.745 + log (0.50/0.75)= 4.57
Is this solution a buffer? Why or why not?
25.0 ml of 0.300 M HNO3 and 50.0 ml of 0.500 M NaC2H3O2.
R HNO3 + NaC2H3O2 ---> NaNO3 + HC2H3O2
I 7.5 mmol 25.0 mmol 0 0
C -7.5 -7.5 +7.5 +7.5
E ~0 17.5 7.5 7.5
The solution contains a ratio of acid/conjugate base that is greater than 0 and less than 100, and therefore is a buffer.
What is the pH of a solution at equilibrium containing 0.15 M NH4+and 1.5 M NH3? The Kb of NH3 is 1.8 x 10-5.
NH3 + H2O --> NH4 + OH- (this is a buffer solution)
pKb =-log(1.8x10-5)= 4.745, pKa= 14-4.745= 9.255
pH = 9.255 + log (1.5/0.15)= 10.26
Calculate the pH of a solution consisting of 15 g of HF (20 g/mol) and 21 g of NaF (42 g/mol). Ka for HF is 7.2 x 10-4.
15 g HF x (1 mol/ 20 g)= 0.75 mol HF
21 g NaF x (1 mol/42 g)= 0.50 mol NaF
pH= -log(7.2 x 10-4) + log (0.50/0.75)= 2.97
If 6.0 mL of 1.0 M HCl is added to 100.0 mL of a buffer solution (0.24 M NH3 and 0.20 M NH4+), what is the resulting pH? Ka = 5.6 x 10-10.
HCl is SA, base in buffer will counteract:
R NH3 + HCl --> NH4+ + Cl-
I 24.0 mmol 6.0 mmol 20.0 mmol N/A
C -6.0 -6.0 +6.0 +6.0
E 18.0 0 26.0 w.c.
pH= -log(5.6 x 10-10) + log (18.0/26.0)= 9.09
Calculate the pH of the solution after adding 10.0 mL of 1.00 M NaOH to a 50.0 mL buffer solution consisting of 0.50 M HF and 0.60 M NaF. Ka of HF is 7.2 x 10-4.
Additionally, calculate the resulting change in pH after the addition of the NaOH.
HF --> F- + H+ (NaOH is SB, so acid in buffer will counteract)
R NaOH + HF ---> NaF + H2O
I 10.0 mmol 25.0 mmol 30.0 mmol lots
C -10.0 -10.0 +10.0
E 0 15.0 40.0
pH= -log(7.2 x 10-4) + log (40.0/15.0)= 3.57 (pH of solution after NaOH is added)
Initial pH of solution= -log(7.2 x 10-4) + log (0.60/0.50)= 3.22
Change in pH= 3.57- 3.22= +0.35