Choose the correct one:
1) 25.00 mL is taken from a solution containing NaOH, NaHCO3 and Na2CO3 individually or as a mixture.
up to the phenolphthalein milestone when analyzed
When 11.00 mL of 0.100 M HCl is taken and a second 25.00 mL is taken and titrated to the bromocresol green turning point, 11.02 mL of HCl is consumed. Find the composition of the solution.
a) NaOH
b) NaHCO3
c) Na2CO3
d) NaOH+ Na2CO3
e) NaHCO3 +Na2CO3
2) H2C2O4 with concentrations of 0.0400 M and 0.200 M
Calculate the pH of a solution containing NaHC2O4.
(Ka1 = 5.60x10-2; Ka2 = 5.42x10-5)
a. 1.01
b. 3.45
c. 4.08
D. 5.67
E. 6.89
3) Containing cadmium (Cd2+) and lead (Pb2+) ions
The titration of a 25.00 mL sample with 0.0040 M EDTA consumes 39.10 mL of EDTA. The cadmium in the sample is then masked by forming Cd(CN)4^-2 with excess KCN. This time 28.00 mL for titration
EDTA is wasted. Cd2+ (112.4 g mol-1) in the sample and
Pb2+ (207.2 g mol-1) concentrations are given below
which one?
Pb2+ (M) / Cd2+ (M)
a. 0.112 /0.044
b. 1.1×10-2 / 4.4×10-3
c. 4.48×10-2/8.8×10-3
D. 4.48×10-3/8.8×10-4
E. 8.8×10-3/4.48×10
I am somewhat confused with the problem because it isn't stated the usual way. Usually I see these as taking a single 25.00 mL sample and titrating with HCl. If I understand this problem correctly, it takes 25.00 mL of a sample and titrates with 11.00 to the phenolphthalein end point. Then you take a SEPARATE 25.00 mL of the sample and titrate with 11.02 mL of HCl to the bromocresol end point. If I understand the problem correctly, here is the solution. You need to know the chemistry involved.
Titration of NaOH is NaOH + HCl ==> NaCl + H20. That will take 11.00 mL for the titration to the phenolphthalein end point and 11 mL to the bromocresol end point for a separate titration. It could be NaOH.
Titration of NaHCO3 is NaHCO3 + HCl --> NaCl + H2O + CO2. Titration to the phenolphthalein end point will be zero and a separate sample to the bromocresol green end point will be 11 mL. Can't be NaHCO3.
Titration of Na2CO3 is Na2CO3 + HCl = NaHCO3 + NaCl to the phenolphthalein end point (11 mL HCl needed) AND
NaHCO3 + HCl ==> NaCl + H2O + CO2 from a separate sample (11 mL needed). Therefore, if it were pure Na2CO3 you would need 11 mL for the phenolphthalein and 22 mL for the bromocresol end point. Can't be Na2CO3.
By the same logic, it can't be a mixture of NaOH and Na2CO3 nor can it be a mixture of NaHCO3 and Na2CO3. If you don't see that I can help.
2). I don't understand problem 2 at all.
3). mols Pb = M EDTA x volume EDTA = 0.0040 x 0.028 = 0.000112 in a 0.025 L sample = 0.000112 mols/0.025 L = 0.00448 M Pb^2+
Total volume for Pb^2+ + Cd^2+ = 39.10 mL
- Volume for Pb^2+ ...................= - 28.00 mL
Volume for Cd^2+ ......................... 11.10 mL
So M Cd^2+ = 0.0111 x 0.0040/0.025 = 0.00178
I don't see any answers that fit this. Check my arithmetic. Check you post.