Consider a titration of 100-mL of .25 M formic acid (Ka-1.8 x 10^-4) with .2 M KOH
What is the pH after 100 mL of base has been added?
Given:
M(HCOOH) = 0.25 M
V(HCOOH) = 100 mL
M(KOH) = 0.2 M
V(KOH) = 100 mL
mol(HCOOH) = M(HCOOH) * V(HCOOH)
mol(HCOOH) = 0.25 M * 100 mL = 25 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.2 M * 100 mL = 20 mmol
We have:
mol(HCOOH) = 25 mmol
mol(KOH) = 20 mmol
20 mmol of both will react
excess HCOOH remaining = 5 mmol
Volume of Solution = 100 + 100 = 200 mL
[HCOOH] = 5 mmol/200 mL = 0.025M
[HCOO-] = 20/200 = 0.1M
They form acidic buffer
acid is HCOOH
conjugate base is HCOO-
Ka = 1.8*10^-4
pKa = - log (Ka)
= - log(1.8*10^-4)
= 3.745
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.745+ log {0.1/2.5*10^-2}
= 4.35
Answer: 4.35
Consider a titration of 100-mL of .25 M formic acid (Ka-1.8 x 10^-4) with .2 M...
Consider a titration of 250 mL 0.15 M acetic acid (Ka = 1.8 x10-5) with 0.10 M KOH. What is the pH of the acetic acid solution (ie: before the titration has begun?) What is the pH after adding 25 mL of 0.10 M KOH? What is the volume of base needed to reach the equivalence point? Is the pH at equivalence point acidic, basic, or exactly neutral? What is the pH after 500 mL of KOH has been added?
Consider the titration of 100.0 mL of 0.200 M acetic acid ( Ka = 1.8 x 10-5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a 0.0 mL pH= 50.0 mL pH = C 100.0 mL pH = 140.0 mL pH= C 200.0 mL pH = f 240.0 mL pH=
Determine the pH during the titration of 25.5 mL of 0.455 M formic acid (Ka = 1.8×10-4) by 0.477 M KOH at the following points. (a) Before the addition of any KOH (b) After the addition of 5.90 mL of KOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 36.5 mL of KOH
2. Consider the titration of 25.0mL of 0.100M formic acid with 0.100 M NaOH (Ka= 1.8 x 10-4). Part IV. What is the pH after adding 25.0mL of NaOH? a)5.60 b) 8.23 c) 9.80 d)13.00 Answer is B
3) (10 points total) A 25.0-mL sample of 0.35 M HCOOH (formic acid) is titrated with 0.20 M KOH. What is the pH of the solution after 25.0 mL of KOH has been added to the acid? Ka-1.77 x 10 a) (4 points) What is the pH of the solution after 25.0 mL of KOH has been added to the acid? Ks-1.77 x 10 C 2-l04Co1s5 b) (6 points) Calculate at least nine (9) more titration points, build a table...
Consider the titration of 50 mL of 0.100 M acetic acid (CH3COOH, Ka = 1.8 x 10-5) with 0.200 M NaOH solution. Show all calculations for full credit. a) Write the titration reaction: b) Calculate the pH after 5.00 mL of NaOH: c) Calculate the pH after 12.5 mL of NaOH: d) Calculate the pH after 25 mL of NaOH:
Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3COOH(aq) with 0.2000 M NaOH(aq) after 2.5 mL of the base have been added. Ka of acetic acid = 1.8 x 10-5. QUESTION 9 Calculate the pH during the titration of 30.00 mL of 0.1000 M ethylamine, C2H5NH2(aq), with 0.1000 M HCl(aq) after 26 mL of the acid have been added. Kb of ethylamine = 6.5 x 10-4. QUESTION 10 Calculate the pH during the titration of 20.00...
25. 50.00 mL of 0.10 M CH3COOH (Ka = 1.8 x 10") is titrated with a 0.10 M KOH solution. After 75.00 mL of the KOH solution is added, the pH in the titration flask will be A) 9.31 B) 9.18 C) 9.52 D) 11.63 E) 12.30
Hydrofluoric acid has a Ka of 6.8 x 10-4. In this titration, 20.00 mL of 0.120 MHF is titrated against 0.080 M KOH. Calculate the pH when 40.00 mL of base has been added. 12.12 O 7.88 12.73 12.41
(ueak acid/shing base 3. Calculate the pH at the following points for the titration of 25.0 mL of 0.100 M formic acid (Ka-1.80 x 10") with 0.100 M NaOH: VNOH 0.00 mL, 15.00 mL, 25.00 mL, 40.00 mL. Draw a graph of pH vs. VaoH. (30 points) (b) Buffer capacity can be thought of as how well a solution resists changes in pH after a strong base/acid is added. A buffer is most effective to resisting pH changes when what...