Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with potassium hydroxide. It requires 22.9 mL of 1.430 M KOH solution to titrate both acidic protons in 54.2 mL of the carbonic acid solution.

Required:
a. Write a balanced net ionic equation for the neutralization reaction. Include physical states.
b. Calculate the molarity of the carbonic acid solution.

Answers

Answer 1

Answer:

a. H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)

b. 0.603 M

Explanation:

Step 1: Write the neutralization reaction

H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)

Step 2: Calculate the reacting moles of KOH

22.9 mL of 1.430 M KOH react.

0.0229 L × (1.430 mol/L) = 0.0327 mol

Step 3: Calculate the reacting moles of H₂CO₃

The molar ratio of H₂CO₃ to KOH is 1:1. The reacting moles of H₂CO₃ are 1/1 × 0.0327 mol = 0.0327 mol.

Step 4: Calculate the molarity of H₂CO₃

0.0327 moles of H₂CO₃ are in a volume of 54.2 mL. The molarity of H₂CO₃ is:

M = 0.0327 mol/0.0542 L = 0.603 M


Related Questions

Solid diarsenic trioxide reacts with fluorine gas (F2) to produce liquid arsenic pentafluoride and oxygen gas (O2). Write the Qc for this reaction.

Answers

Answer:

QC= [O2]^3/[F2]^10

Explanation:


1. What 2 subatomic particles have charges? List the particle name and its charge.

Answers

Answer: Proton - positive charge (+)

Neutron - neutral charge (0)

Electron - negative charge (-)

Explanation:

In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.

Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?

Answers

Answer:

a) - 0.2 M

b) - 0.2 M

c)- 0

Explanation:

The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:

MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol

a). Molarity = moles CuBr₂/1 L solution

moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol

Volume in L = 375 mL x 1 L/1000 mL = 0.375 L

M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M

b). When is added to water, CuBr₂ dissociates into ions as follows:

CuBr₂ ⇒ Cu²⁺ + 2 Br⁻

We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:

0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M

c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).

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