A 25.0 mL sample of 0.150 M formic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of formic acid is 1.8 ⋅ 10-4.
1 mol of NaOH reacts with 1mol of formic acid
Molarity of Formic acid (M1) = 0.150 M
Volume of formic acid (V1) = 25.0 mL
Molarity of NaOH (M2) = 0.150 M
Let the volume of NaOH required at equivalence point be V2
From Molarity equation
M1V1 = M2V2
V2 = 0.150 M * 25.0 mL / 0.150 M = 25 mL
Total volume of solution = 25 + 25 = 50.0 mL = 0.075 L
[HCOO-] = Molarity *volume = 0.150M * 0.025 L/0.050L = 0.075 moles
Kb = Kw/Ka = (10-14 ) /(1.8 *10-4) = 5.56 *10-11
HCOO- + H2O <--> HCOOH + OH-
Kb = [HCOOH][OH-] / [HCOO-]
5.56 *10-11 = x2 / (0.075)
x = 2.04 *10-6 M = [OH-]
pOH = -log [OH-] = -log (2.04 *10-6 M) = 5.69
pH = 14 - pOH = 14 -5.69 = 8.31
A 25.0 mL sample of 0.150 M formic acid is titrated with a 0.150 M NaOH...
A 25.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of acetic acid is 4.5E-4
A 25.0 mL sample of 0.150 M chloroacetic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The of chloroacetic acid is 1.4 ×10-3.
A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH after the addition of 13.0 mL of NaOH? The Ka of benzoic acid is 6.3x10-5.
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...
8) A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH at the eq. point? pka= 4.20
A 50.0 mL sample of 0.25 M formic acid (HCOOH) aqueous solution is titrated with 0.125 M NaOH solution. Ka of HCOOH = 1.7 x 10−4. a. Calculate the pH of the solution after 50 mL of NaOH solution has been added. b. How many mL of 0.125 M NaOH need to be added to the sample to reach the equivalence point? What is the pH at the equivalence point?
A 50.0 ml sample of 0.50 M acetic acid, ch3cooh is titrated with a 0.150 M NaOH solution. calculate the ph after 25.0 ml of the base have been added (ka=1.8x10^-5)
25. A 50.0 mL sample of 0.150 M weak acid was titrated with a 0,150 M NaOH solution. What is the pH after 30.0 mL of the sodium hydroxide solution is added? The Ka of the acid is 1.9x10(3 points) D) 4.78 E) None of these C) 3.03 (A) 4.90 B) 1.34 26. A 25.0 mL sample of 0.25 M hydrofluoric acid (HF) is titrated with a 0.25 M NaOH solution. What is the pH after 38.0 mL of base...
A 25.0 mL sample of 0.293 M acetic acid solution is titrated with 37.5 mL of 0.195 M NaOH solution. What is the pH of this solution if K = 1.8 x 10-5? Final Answer: _______
A student titrated a 100.0 mL sample of 0.100 M acetic acid with 0.050 M NaOH. (For acetic acid, Ka = 1.8 * 10^-5 at this temperature.) (a) Calculate the initial pH. (b) Calculate the pH after 50.0 mL of NaOH has been added. (c) Determine the volume of added base required to reach the equivalence point. (d) Determine the pH at the equivalence point?