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Fofino [41]
3 years ago
13

If 11.00 mL of a standard 0.2831 M NaOH solution is required to neutralize 26.86 mL of H2SO4, what is the molarity of the acid s

olution?
Chemistry
1 answer:
MrRa [10]3 years ago
3 0

Answer:

0.1159 M

Explanation:

Using the formula below:

CaVa = CbVb

Where;

Ca = concentration/molarity of acid (M)

Cb = concentration/molarity of base (M)

Va = Volume of acid (mL)

Vb = volume of base (mL)

According to this question, the following information were given:

Ca = ?

Cb = 0.2831 M

Va = 26.86 mL

Vb = 11.00 mL

Using CaVa = CbVb

Ca × 26.86 = 0.2831 × 11

26.86Ca = 3.1141

Ca = 3.1141 ÷ 26.86

Ca = 0.1159

The molarity of the acid (H2SO4) solution is 0.1159 M

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Answer:

A. 0.000128 M is the solubility of M(OH)2 in pure water.

B. 3.23\times 10^{-6} M is the solubility of M(OH)_2 in a 0.202 M solution of M(NO_3)_2.

Explanation:

A

Solubility product of generic metal hydroxide = K_{sp}=8.45\times 10^{-12}

M(OH)_2\rightleftharpoons M^{2+}+2OH^-

                      S         2S

The expression of a solubility product is given by :

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0.000128 M is the solubility of M(OH)2 in pure water

B

Concentration of M(NO_3)_2 = 0.202 M

Solubility product of generic metal hydroxide = K_{sp}=8.45\times 10^{-12}

M(OH)_2\rightleftharpoons M^{2+}+2OH^-

                   S          2S

So, [M^{2+}]=0.202 M+S

The expression of a solubility product is given by :

K_{sp}=[M^{2+}][OH^-]^2

8.45\times 10^{-12}=(0.202 M+S)(2S)^2

Solving for S:

S=3.23\times 10^{-6} M

3.23\times 10^{-6} M is the solubility of M(OH)_2 in a 0.202 M solution of M(NO_3)_2.

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