1. Chromic acid is a diprotic acid H 2 CrO 4 . A solution of potassium chromate

Question

1. Chromic acid is a diprotic acid H2CrO4. A solution of potassium chromate is prepared at pH 5.97. Calculate the fraction of the total chromate in this solution that is present as the chromate ion CrO42-.  Use Ka data given in Harris 8th edition.

2. Calculate the molar solubility of Ag2CrO4 at pH 5.57.  Use Ksp for Ag2CrO4 and Ka data for chromic acid.

3. A 1.2614-g sample of primary standard CaCO3 was dissolved in 1:1 HCl and diluted to exactly 1 liter. Calculate the molarity of the resulting CaCl2 solution.

4. A 25.00-mL aliquot of 0.01122 M CaCl2 solution required 23.43 mL of EDTA for titration at pH 10 using calmagite as indicator. Calculate the molarity of the EDTA solution.

5. A 250.0-mL sample of Lake Superior water was adjusted to pH 10 with ammonia buffer and calmagite indicator was added. The prepared sample required 25.89 mL of 0.00891 M EDTA for titration to the endpoint. Calculate the hardness as ppm CaCO3.

6.  At pH 12, magnesium hydroxide is insoluble and does not react with EDTA. The titration of a hard water sample at pH 12 thus yields results for calcium only. A 250.0-mL sample of tap water was adjusted to pH 12 and titrated with 0.01 EDTA using calcein indicator, requiring 20 mL to reach the endpoint. A 200.0-mL sample of the same tap water required 48 mL of the standard EDTA for titration at pH 10 using calmagite indicator.

Calculate the molar concentration of Mg2+ in the tap water.

7. Calculate pCu at the equivalence point in the titration of 20.0 mL of 0.011700 M CuSO4 with 0.0100 M EDTA at pH 6.00. Use Kf for Cu-EDTA and alpha for EDTA.

Explanation / Answer

1) Ka = 1.2*10^-6

H2CrO4 ------> 2H+ + CrO4-2 ; [H+] = 10^-pH = 1.07*10^-6 M

[CrO4-2]/[H2CrO4] = Ka * [H+]^2 = 1.38*10^-18

2) Ag2CrO4 + 2H+ ---> 2Ag+ + H2CrO4 ; Keq = Ksp / Ka ; [H+] = 10^-pH =

[Ag+] = Keq *[H+]^2 = 1.2*10^-17 M {Ksp = 2*10^-12}

3) Moles of Calcium = moles of CaCO3 = mass/molar wt = 0.012614 mol

Moles of CaCl2 formed = moles of Ca = 0.012614

Hence, Molarity = Moles/Vol = 1.26*10^-2 M

4) V1M1 = V2 M2

M2 = 25*0.01122 / 23.43 = 0.012 M

5) Moalrity of water = 25.89*0.00891 / 250 = 9.227 * 10^-4 moles Ca++ ion / L

ppm is mg/L

9.227 * 10^-4 moles Ca++ ion / L = 9.227 * 10^-4 * 100 g CaCO3 ion / L

= 9.227 * 10^-4 *100*1000 mg CaCO3 ion / L = 92.27 ppm

6) Molarity of Ca++ (Use 1st titration) = V1M1/V2 = 8*10^-4 moles/L

Molarity of Ca+2 and Mg+2 (use 2nd titration) = 2.4*10^-3 moles / L

So moles of Mg+2 per L = 1.6*10^-3 moles

Conc = 1.6*10^-3 M

7) I could not get the values on9. Sorry, could not solve it. Also, explain pCu..??

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