Which is the percent ionization of a solution which is 0.222 M HNO2 (K a = 4.50

Question

Which is the percent ionization of a solution which is 0.222 M HNO2 (Ka = 4.50 x 10-4) and 0.278 M KNO2?

A. 0.02%

B. 0.162%

C. 1.55 %

D. 3.45 %

A. 2.06

B. 5.48

C. 8.52

D. 9.26

Which process best accounts for the numerical value of the equivalence-point pH after the titration of formic acid, HCOOH with NaOH?

A. HCOO- (aq) + H2O (l) <--------> HCOOH (aq) + OH- (aq)

B. HCOOH(aq) + H2O(l) <------->   -COOH (aq) + H3O+ (aq)

C. HCOOH (aq) + H3O+ (aq) <--------> HCOOH2+ (aq) + H2O(l)

D. HCOOH(aq) + H2O(l) <-------> HCOOH2+ (aq) + OH- (aq)

Which volume of 0.234 M HCl must be added to 45.0 mL 0.432 M CH3NH2 (Kb = 3.70 x 10-4) to reach a pH of 11.21?

A.   5.00 mL

B. 11.6 mL

C. 14.0 mL

D. 83.0 mL

A. 0.02%

B. 0.162%

C. 1.55 %

D. 3.45 %

Which is the equivalence-point pH if 25.0 mL HCOOH (1.8 x 10-4) requires 29.80 mL 0.3567 M NaOH to reach the equivalence point?
Answer

A. 2.06

B. 5.48

C. 8.52

D. 9.26

1. Which process best accounts for the numerical value of the equivalence-point pH after the titration of formic acid, HCOOH with NaOH?

   Answer

   A. HCOO- (aq) + H2O (l) <--------> HCOOH (aq) + OH- (aq)

   B. HCOOH(aq) + H2O(l) <------->   -COOH (aq) + H3O+ (aq)

   C. HCOOH (aq) + H3O+ (aq) <--------> HCOOH2+ (aq) + H2O(l)

   D. HCOOH(aq) + H2O(l) <-------> HCOOH2+ (aq) + OH- (aq)

   Which volume of 0.234 M HCl must be added to 45.0 mL 0.432 M CH3NH2 (Kb = 3.70 x 10-4) to reach a pH of 11.21?

   Answer

   A.   5.00 mL

   B. 11.6 mL

   C. 14.0 mL

   D. 83.0 mL

Explanation / Answer

HNO2 ----> H+ + NO2-

0.222-x x 0.278+x

Ka = x*(0.278+x)/(0.222-x) = 4.5*10^-4

x = 0.000358 M = [H+]

so percent ionisation = [H+]/[HNO2] *100 = 0.162%........ansB

ans2)

HCOOH + NaOH %u2192 HCOO- + Na+ + H2O

moles of [NaOH] = 29.8*0.3567 = 10.62966 m moles

so [HCOO-] = 10.62996/(29.8+25) = 0.1939 M

also Kb = Kw/Ka = 5.55*10^-11

HCOOH + NaOH %u2192 HCOO- + Na+ + H2O

HCOO- + H2O ----> HCOOH + OH-

0.1939-x x x

Kb = x^2/(0.1939-x)

x = 3.28*10^-6 = [OH-]

pH = 14+log10[OH-] = 8.515 = 8.52 ............ANSC

ans 3)

A. HCOO- (aq) + H2O (l) <--------> HCOOH (aq) + OH- (aq)

A. HCOO- (aq) + H2O (l) <--------> HCOOH (aq) + OH- (aq)

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