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Paladinen [302]
3 years ago
6

Calculate the concentration of acetate ion in a buffer solution made from 2.00 mL of 0.50 M acetic acid and 8.00 mL of 0.50 sodi

um acetate .
Chemistry
1 answer:
Lelu [443]3 years ago
8 0

Answer:

1 M

Explanation:

Equation of reaction is;

CH3COOH + CH3COONa -------------------> 2CH3COO^- + NaH

1 moles of each of the reactants react to give 2 moles of the acetate ion.

From the question, we have that 2.00 mL that is (2÷1000)L of 0.50 M acetic acid reacted with 8.00 mL that is (8/1000)L of 0.50 sodium acetate.

Then from equation, n = CV -------------------------------------------(1).

Where n= number of moles, V= volume, C= concentration.

Number of moles,n of acetic acid = 0.50M× 2/1000L.

n(acetic acid)= 0.001 moles.

Number of moles,n of sodium acetate= 0.50M ×(8/1000)L.

n(sodium acetate)= 0.004 moles.

0.001 moles of acetic acid react with 0.004 moles of Sodium acetate

Therefore, acetic acid is the limiting reagent.

One mole of acetic acid produces 2 moles of acetate ion.

0.001 mole of acetic acid produces= 0.002 moles of acetate ion.

Using the equation (1) that is, n= CV.

0.002= C× 2/1000

C= 0.002/0.002

C= 1 M

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Read 2 more answers
Which is the limiting reagent in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2
NemiM [27]

Answer:

CuSO4

Explanation:

Na2S + CuSO4 → Na2SO4 + CuS

The reaction is balanced (same number of elements in each side)

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*Remember mole = mass / MM

With that information now you have to divide each moles by its respective stoichiometric coefficient

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CuSO4 stoichiometric coefficient: 1

CuSO4: 0.076 / 1 = 0.076

The smaller number between them its the limiting reagent, CuSO4

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3 years ago
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