7. Calculate the pH of the titration solution after the addition of 49.00 mL of

Question

7. Calculate the pH of the titration solution after the addition of 49.00 mL of 0. 1000 M HCI to 100.0 mL of 0.05000 M solution of NaOH. (A) 10.83 (B) 7.00 (C) 2.70 (D) 1.00 (E) 12.70 8. 50.00 mL of 0.1000 M I- is titrated with 0.1000 M AgNO, What is the pl (pl--log 1UDf the solution at the equivalence point of the titration? The of Agl is 8.3x10 (A) 1.40x10 (B) 8.04 (C) 3.16 (D) 7.00 (E) 1.00 9 Back titration is a procedure in which titrant is added to analyte until the reaction is complete. the weight of the titrant is measured instead of its volume. excess standard reagent is added to the analyte and the former is titrated with a second standard reagent the reaction product is titrated with a second standard reagent excess analyte is titrated with a second standard reagent. (A) (B) (C) (D) (E) 10. The charge balance for an aqueous solution containing HO.H.. OH, Ca". K. CHOH CrO2, and Ch is: (B) H0]+[H+ICa]+ (K1+ (C H.OH] IOH IC+ICro*1

Explanation / Answer

Q7

mmol of HCl = MV = 49**0.1 = 4.9

mmol of NaOH = MV = 0.05*100 = 5

therefore, excess base

mmol of NAOH left = 5-4.9 = 0.1

Vtotal = 100+49 = 149 mL

[NaOH] =) mmol/mL = 0.1/149 = 0.0006711 M

pOH = -log(0.0006711 = 3.1732

pH = 14-pOH = 14-3.1732

pH = 10.8268

Q8

in equivalence point

mmol of I- = MV = 0.1*50 = 5

mmol of Ag required = 5

V AgNO3 = mmol/M = 5/0.1 = 50 mL

Vtotal = 50+50 = 100 mL

in equilbirium

Ksp = [Ag+][I-]

(8.3*10^-17) = S*S

S = sqrt((8.3*10^-17) ) = 9.110*10^-9

pI = -log(S) = -log( 9.110*10^-9) = 8.04

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.