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 of 0.25 M of formic acid = 50 x 0.25 = 12.5 mol formic acid
50.0 mL of 0.125 M of NaOH solution = 50 x 0.125 = 6.25 mol NaOH
6.25 mol NaOH will react with 6.25 mol of HCOOH to give 6.26 mol of HCOONa.
Therefore, the moles of remaining HCOOH = (12.5 - 6.25) = 6.25 mol
Total volume of the solution = (50 + 50) = 100 mL
Concentration of HCOOH = 6.25/100 = 0.0625 M
Concentration of HCOONa = 6.25/100 = 0.0625 M
Ka of HCOOH = 1.7 x 10−4
From Henderson-Hasselbalch equation,
pH = pKa + log [base]/[acid]
= - log(1.7 x 10−4) + log 0.0625/0.0625
= 3.77
-----
b.
Volume of formic acid, V1 = 50.0 mL
Concentration of formic acid, S1 = 0.25 M
Volume of NaOH = V2
Concentration of NaOH, S2 = 0.125 M
Now, equating the moles of acid and the base we get,
V1S1 = V2S2
or, 50.0 x 0.25 = V2 x 0.125
or, V2 = 100 mL
Therefore, the volume of the NaOH solution need to be added = 100 mL
---
Moles of HCOONa forms at equivalence point = 100 x 0.125 = 12.5 mol
Total volume of the solution = 100 + 50 = 150 mL
Concentration of HCOONa = 12.5/150 = 0.083 M
Ka of HCOOH = 1.7 x 10−4
Therefore, Kb of HCOO- = 10-14/(1.7 x 10−4) = 5.88 x 10-11
HCOONa + H2O HCOOH + OH-
Initial concentration (M) 0.083 0 0
Change in concentration (M) -X X X
Equilibrium concentration (M) (0.083 - X) X X
Now,
Kb = [HCOOH][OH-]/[HCOONa]
or, 5.88 x 10-11 = X2/(0.083 - X) [(0.083 - X) 0.083 as X << 0.083]
or, 5.88 x 10-11 = X2/0.083
or, X2 = 5.88 x 10-12
or, X = 2.21 x 10-6
or, [OH-] = 2.21 x 10-6
or, - log[OH-] = - log(2.21 x 10-6) [Taking -log on both sides of the equation]
or, pOH = 5.66
or, 14 - pH = 5.66 [As pH + pOH = 14]
or, pH = 8.34
Therefore, the pH at the equivalence point = 8.34
A 50.0 mL sample of 0.25 M formic acid (HCOOH) aqueous solution is titrated with 0.125...
a 100 ml solution of 0.250 M formic acid (HCOOH) was titrated to its equivalence point with 50 mL of sodium hydroxide. The complete molecular equation for the reaction is shown below HCOOH (aq) + NaOH (aq)---------> HCOONa (aq) +H20 (l) Ka of HCOOH= 1.7 x 10 ^-4 calculate the pH at the equivalence point
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...
A 50.0-mL sample of 0.15 M butanoic acid, CH3CH2CH2COOH, is titrated with 0.30 M NaOH(aq). Ka for butanoic acid is 1.52 x 10-5 a) How many mL of NaOH(aq) are required to reach the equivalence point? b)What is the pH of the solution after 27.0 mL of NaOH(aq) have been added?
A 50.0 mL sample of a 0.517 M aqueous hypochlorous acid solution is titrated with a 0.206 M aqueous barium hydroxide solution. What is the pH after 40.7 mL of base have been added? pH =
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?
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 0.150 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.
What is the pH of the solution when 100 mL of 0.8 M formic acid (Ka= 1.8x10-4) is titrated with 50 mL of 1.2 M NaOH? 1.80 grams of an unknown monoprotic acid (HA) required 48.62 mL of a 0.25 M NaOH solution to reach the equivalence point. Calculate the molar mass of the acid. Need help understanding please show work. Thank you in advance.
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?
When a 23.8 mL sample of a 0.443 M aqueous hydrofluorie acid solution is titrated with a 0.358 M aqueous barium hydroxide solution, what is the pH after 22.1 mL of barium hydroxide have been added? pH- What is the pH at the equivalence point in the titration of a 19.7 mL sample of a 0.376 M aqueous nitrous acid solution with a 0.447 M aqueous sodium hydroxide solution? pH- When a 20.1 mL sample of a 0.417 M aqueous...