Of 0.00MNaC t 500m 8. Will the precipitation of r be complete when you add 50. 0
ID: 1040564 • Letter: O
Question
Of 0.00MNaC t 500m 8. Will the precipitation of r be complete when you add 50. 0.040 M AgNO3? 9. Will the precipitation of Ca2 as CaC,09 be complete if (Ca] 0.0o50 M and [C,o1 0.0100 M are in solution. Assume 1.00 L total volume. Kip 2.7 x 109. Step 1: Does precipitation occur? Step 2: How much Ca2+ left after reaction and at equilibrium? er What is the pH of the solution at equilibrium if 1.00 mol of acetic acid and 0.5o0 mol of sodium acetate are mixed together in 1.00L of water? Can this mixture be considered a buffer? 1.00 M CH,COOH/O.500 M CH,COO. 0.10 mol of hydrochloric acid are added to this buffer solution with no significant change in volume. What is the pH of the resulting solution at equilibrium? Should it be close to the original pH? What change will be caused by the addition of a small amount of HCI to a HE/NaF buffer? J3. Design à buffer with a pH of 4.60 14. Calculate the pH of a solution prepared by mixing 100.0ml of 1.20M ethanolamine, CaH,ONH, with 0.0ml of 1.0M HCI. K, for CH,ONH' is 3.61x100 15. What is the pH of a solution that is formed by titrating 450 mL of o.205 M NaOH with 740 mL of 0.012 M HCI? Is this solution at its equivalence point 16. Hypochlorous acid (HOCI) is a weak acid with pk,-7.54. Suppose a 45.00 mL sample of 0.240 M HOCI solution is titrated with a 0.250 M standard OH solution. What is the pH at the equivalence point? 17. Calculate the [CH,COO] if the solution consists of 0.010 M HCI and 0.015 M CH,COOH.Explanation / Answer
9) Qsp of CaC2O4 = [Ca2+][c2o4-2]
= 0.005*0.01
= 5*10^-5
Qsp > Ksp . so that, precipitation takes place.
ste2. Ksp = S*S
(2.7*10^-9) = s*s
S = solubility of CaC2O4 = 5.2*10^-5 M
15) no of mol of NaOH = V*M = 450*0.205 = 92.25 mmol
no of mol of HCl = V*M = 740*0.012 = 8.88 mmol
concentration of excess NaOH = (92.25-8.88)/(740+450) = 0.07 M
pH = 14 - (-log(OH-))
= 14 - (-log0.07)
= 12.8
16) at equivalence point,
No of mol of HOCl = no of mol of OH-
No of mol of HOCl = 45*0.24 = 10.8 mmol
No of mol of OH- = 10.8 mmol
VOLUME of OH- must add = n/M = 10.8/0.25 = 43.2 ml
concentration of clO- base = 10.8/(45+43.2) = 0.122 M
pH = 7+1/2(pka+logC)
= 7+1/2(7.54+log0.122)
= 10.31
17) CH3COOH <-----> CH3COO-(aq) + H+(aq)
HCl --------> Cl-(aq) + H+(aq)
Ka = [CH3COO-][H+]/[CH3COO-]
(1.8*10^-5) = (x*(0.01+x))/(0.015-x)
x = 2.69*10^-5 M
[ch3coo- ] = 2.69*10^-5 M
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