A 41.0 mL sample of 0.174 M HNO2 is titrated with 0.264 M KOH. (Ka for HNO2 is 4.57×10−4.)
Determine the pH at the equivalence point for the titration of HNO2 and KOH.
find the volume of KOH used to reach equivalence point
M(HNO2)*V(HNO2) =M(KOH)*V(KOH)
0.174 M *41.0 mL = 0.264M *V(KOH)
V(KOH) = 27.0227 mL
Given:
M(HNO2) = 0.174 M
V(HNO2) = 41 mL
M(KOH) = 0.264 M
V(KOH) = 27.0227 mL
mol(HNO2) = M(HNO2) * V(HNO2)
mol(HNO2) = 0.174 M * 41 mL = 7.134 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.264 M * 27.0227 mL = 7.134 mmol
We have:
mol(HNO2) = 7.134 mmol
mol(KOH) = 7.134 mmol
7.134 mmol of both will react to form NO2- and H2O
NO2- here is strong base
NO2- formed = 7.134 mmol
Volume of Solution = 41 + 27.0227 = 68.0227 mL
Kb of NO2- = Kw/Ka = 1*10^-14/4.57*10^-4 = 2.188*10^-11
concentration ofNO2-,c = 7.134 mmol/68.0227 mL = 0.1049M
NO2- dissociates as
NO2- + H2O -----> HNO2 + OH-
0.1049 0 0
0.1049-x x x
Kb = [HNO2][OH-]/[NO2-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((2.188*10^-11)*0.1049) = 1.515*10^-6
since c is much greater than x, our assumption is correct
so, x = 1.515*10^-6 M
[OH-] = x = 1.515*10^-6 M
use:
pOH = -log [OH-]
= -log (1.515*10^-6)
= 5.8196
use:
PH = 14 - pOH
= 14 - 5.8196
= 8.1804
Answer: 8.18
A 41.0 mL sample of 0.174 M HNO2 is titrated with 0.264 M KOH. (Ka for...
a) A 41.0 mL sample of 0.194 M HNO2 is titrated with 0.220 M KOH. (Ka for HNO2 is 4.57×10−4.) Determine the pH at the equivalence point for the titration of HNO2 and KOH. b) A 50.0-mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M nitric acid. Calculate the pH of the solution, after you add a total of 56.7 mL 0.200 M HNO3.
A 42.0 mL sample of 0.120 M HNO2 is titrated with 0.214 M KOH. (Ka for HNO2 is 4.57×10−4.) Determine the pH at the equivalence point for the titration of HNO2 and KOH.
A 40.0 mL sample of 0.150 M HNO2 (Ka = 4.60 x 10-4) is titrated with 0.200 M KOH. Calculate: a. the pH after adding 10.00 mL of KOH b. the pH at one-half the equivalence point c. the pH after adding 20.00 mL of KOH d. the volume required to reach the equivalence point e. the pH at the equivalence point f. the pH after adding 45.00 mL of KOH
Having trouble figuring out this titration problem in my homework! I appreciate any help with solving it! Thank you in advance! A 41.0 mL sample of 0.158 M HNO2 is titrated with 0.202 M KOH. (Ka for HNO2 is 4.57×10−4.) Determine the pH at the equivalence point for the titration of HNO2 and KOH.
A 15 mL sample of hydrofluoric acid, Ka=6.8x10^-4 was titrated with .475 M KOH. The equivalence point was reached after addition of 19.35 mL of base. Determine the molar concentration of the original hydrofluoric acid solution, and find the pH of the solution.
A 35.0 mL sample of 0.136 M HNO, is titrated with 0.230 M KOH (K, for HNO, is 4.57x10-4) Part A Determine the pH at the equivalence point for the titration of HNO, and KOH V AED pH = Submit Request Answer Provide Feedback Type here to search
9. 100 mL of 0.5M HNO2 (Ka = 4.5*10-4) is titrated with 0.5M KOH. What is the pH after 100mL of 0.5M KOH is titrated into the acid? (12)
50.0 mL of 0.090 M nitrous acid (HNO2, Ka = 7.1 x 10-4), is titrated with 0.100 M NaOH, requiring 45.0 mL of strong base to reach the equivalence point. (a) What will be the pH after 35.0 mL of NaOH have been added? (b) What will be the pH at the equivalence point? (c) What will be the pH after 60.0 mL of NaOH have been added?
1.A 25.00 ml smaple of 0.523 M nitrous acid, HNO2, solution is titrated with a 0.213 M NaOH. For HNO2, Ka = 4.0 X 10^-4 a) What is the pH before any NaOH is added? b) Write the reaction that takes place as KOH solution is added to the HNO2 solution. c) write the reaction that determines the pH at the equivialnce point. -What is the pH at the equivilance point -what is the pH at the 1/2 equivilance point...
50.0 mL of 0.275 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point? Ka for HNO2 is 4.0x10^-4