2. ACID Problem Calculate the pH of a solution of 103 M chromic acid (H2CrOa): p
ID: 1039868 • Letter: 2
Question
2. ACID Problem Calculate the pH of a solution of 103 M chromic acid (H2CrOa): pKa1 0.74; pKa2 6.49 both at 25*C 3. Buffer1 Calculate the pH of a buffer solution prepared by mixing 10 g KH2PO4 with 50 g K2HPOs in 1 liter of water. 4. Buffer2 Find a buffer to maintain a laboratory system for coagulation at a pH of 5.5. The system uses a ferric chloride coagulant and generates acidity as defined by the reaction in equation (1). If the coagulant dose is 10 mg/L Fe3 calculate the concentration of the buffer to maintain the pH at a level within 0.2 pH units. FeCla +3 H20Fe(OH)s (s)+3H + 3Ch 5. ALKALINITY A city water sample has 20 mg/L HCOs and 40 mg/L CO32. What is the alkalinity of this water?Explanation / Answer
2)
H2CrO4(aq) < ---------- > HCrO4-(aq) + H3O+(aq) pka1= 0.74
HCrO4- (aq) < ----------> CrO4^2- (aq) + H3O+(aq) pKa2= 6.49
pKa1 = 0.74
Ka1 = 0.182
Ka1 = [H+][HCrO4-]/[H2CrO4]
= (X × X )/ (10^-3 - X)
0.182 = X^2/(10^-3-X)
X^2 - 0.182X + 1.82×10^-4=0
Solve for X
Then , X = [H+]
pH= - log[H+]
pKa2= 6.49
Ka2 = 3.23×10^-7
Ka2 is very less, So no need for [H+] due to Ka2
3)
Molar mass of KH2PO4 = 136.1 g/mol
Molar mass of K2HPO4 = 174.2 g/mol
[KH2PO4] = mass /(molar mass * volume)
= 10 / (136.1 * 1)
= 0.0735 M
[K2HPO4] = mass /(molar mass * volume)
= 50 / (174.2 * 1)
= 0.287 M
pKa of H2PO4- = 7.21
USE:
pH = pKa + log {[K2HPO4]/[KH2PO4]}
= 7.21 + log (0.287/0.0735)
= 7.80
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