You have titrated 25.0 mL of acid A with 65.35 mL of a 0.231 molar solution of base B. The mole ratio is 1 A = 1 B. What is the concentration of the solution of A?

Question

Question

The answer is 0.604,but i don't know the process

Chemical

Answer 1

Answer #1

a. no of moles = molarity * volume in liters

for NaOH => 0.10 M * 0.013 L => 0.0013 moles

b. acid and the base are in 1:1 ratio means

no of moles are same so the moles of acid is 0.0013 moles

c. using the dilution law

M1V1 = M2V2

for NaOH => M1 = 0.10 M , V1 = 13 mL

for Acid => M2 = ? , V2 = 25 mL

M2 = M1V1 / V2

M2 = 0.10 * 13 / 25

M2 = 0.052 M

molarity of acid is 0.052 M

answered by: ANURANJAN SARSAM

Answer 2

Similar Homework Help Questions

a. A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with...

a. A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with 0.0919 M NaOH and the equivalence point volume was determined by graphical means to be 25.91 mL. What is the concentration of the acetic acid? concentration of HA = ___ M b. Calculate the pKa of a weak monoprotic acid if the pH of a 0.162 M solution is 2.52. pKa = ___
A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with...

A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with 0.1007 M NaOH and the equivalence point volume was determined by graphical means to be 25.42 mL. What is the concentration of the acetic acid? concentration of HA = ___ M
1) A 25.0 mL sample of 0.150 M hydrazoic acid is titrated with a 0.150 M...

1) A 25.0 mL sample of 0.150 M hydrazoic acid is titrated with a 0.150 M NaOH solution. What is the pH after 26.0 mL of base is added? The of hydrazoic acid is 1.9 × 10-5. 2)A 25.0 mL sample of 0.723 M HClO4 is titrated with a KOH solution. The H3O+ concentration after the addition of of KOH is ________ M.

A 25.0-mL sample of 0.10 M weak base is titrated with 0.15 M strong acid. What...

A 25.0-mL sample of 0.10 M weak base is titrated with 0.15 M strong acid. What is the pH of the solution after 9.00 mL of acid have been added to the weak base? Weak base Kb = 6.5 × 10–4
A 25.0 mL sample of an acetic acid solution (Ka = 1.76 × 10-5) is titrated...

A 25.0 mL sample of an acetic acid solution (Ka = 1.76 × 10-5) is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid is __________ M.
Acid/Base titrations 10. 25.0 mL of 0.100 M H2A (a weak diprotic acid) is titrated with...

Acid/Base titrations 10. 25.0 mL of 0.100 M H2A (a weak diprotic acid) is titrated with 0.200 M NaOH. What is the pH of the solution when 0.00 mL, 10.0 mL, 12.5 mL, 20.0 mL, 25.0 mL, and 40.0 mL E.S RaWeMA 5.83 x 10 8) have been added? (Ka1= 2.46 x 10, Ka2 ANSWER: 0 mL = 2.315, 10.0 mL= 4.211 (or 4.213), 12.5 mL = 5.422, 20.0 mL 7.410, 25.0 mL = 9.966, 40.0 mL = 12.664 11....

A 25.0 mL of a weak acid is titrated with a strong base (0.1 M). Calculate...

A 25.0 mL of a weak acid is titrated with a strong base (0.1 M). Calculate the pH of the solution during the titration if the weak acid concentration is 0.10 M and its Ka = 1.8 x 10-5 and 10.0 mL of base has been added. (Hint: use Henderson-Hasselbach equation). Question options: a) pH= 7.00 b) pH= 5.28 c) pH = 4.56 d) pH= 4.74
B. Molar Concentration of an Acid Solution Acid type, HA or H,A: HAUnknown No. Balanced equation...

B. Molar Concentration of an Acid Solution Acid type, HA or H,A: HAUnknown No. Balanced equation for neutralization of acid with NaOH Sample 3 25.0 24.0 Sample Sample 2 1. Volume of acid solution (ml) 2. Buret reading of NaOH, initial (mL) 3. Buret reading of NaOH. final (ml) 4. Volume of NaOH dispensed (mL) 5. Molar concentration of NaOH (mol/L), Part A 6. Moles of NaOH dispensed (mol) 7. Molar concentration of acid solution (mol/L) 25.0 25.0 IH. 8....
Calculate the pH of a 25.0 mL of 0.100M base acetic acid solution after being titrated...

Calculate the pH of a 25.0 mL of 0.100M base acetic acid solution after being titrated with 0.100 M NaOH to its equivalence point (pKb (acetic acid)=5.68x10^-10)

5-10 molar concentration of NaOH solution is 0. 128 B. Molar Concentration of an Acid Solution Acid type HA HA H...

5-10 molar concentration of NaOH solution is 0. 128 B. Molar Concentration of an Acid Solution Acid type HA HA HA Unknown No. Balanced equation for beutralization of acid with NaOH. 25.0 25.0 25.0 1. Volume of acid solution (m) 2 Buret reading of NaOH, initial (L.) 3. Buret reading of NaOH,final (mL) 4. Volume of NaOH dispensed (m) 5. Molar concentration of NaOH (MOVL). Part A 6. Moles of NaOH dispensed (mol) 7. Molar concentration of acid solution (mol/L)...