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shepuryov [24]
3 years ago
7

In a titration experiment, 13.0 mL of an aqueous HCl solution was titrated with 0.2 M NaOH solution. The equivalence point in th

e titration was reached when 11.1 mL of the NaOH solution was added. What is the molarity of the HCl solution? HCl molarity
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
8 0

Answer:

The molarity of the HCl solution is 0.17M

Explanation:

When a titration occurs between an acid and a base it is called neutralization.

The formula to calculate the concentration of the acid is:

Ca*Va=Cb*Vb

where Ca and Cb are the concentrations of the acid and the base respectively and Va and Vb are the volumes of the acid and the base.

As HCl is an acid, solve the equation for Ca:

Ca=\frac{Cb*Vb}{Va}

Replacing values:

Ca=\frac{0.2MNaOH*11.1mLNaOH}{13.0mLHCl}

Ca=0.17M

Therefore the molarity of the HCl solution is 0.17M

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An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
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Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

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0=[tex]\delta Q- \delta W

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Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]
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<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq) + Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
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Kb(CN</span>⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻] / [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] = x.
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2.04·10</span>⁻⁵<span> = x² / (0.021 M - x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>

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