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vovangra [49]
2 years ago
12

Determine the pH at the equivalence point for the titration of 40.0 mL of 0.1 M HNO2 by 0.200 M KOH (Example 16.7). (The pKa of

nitrous acid is 3.36.)
Chemistry
1 answer:
Fynjy0 [20]2 years ago
8 0

Answer:

pH=13.28

Explanation:

Nitrous acid

HNO_2 \longrightarrow H^+ + NO_2^-

Ka=\frac{[H^+][NO_2^-]}{[HNO_2]}

pKa=-log(Ka)=3.36

Ka=4.36*10^{-4}

<u>Balance: </u>

[H^+]=X

[NO_2^-]=X

[HNO_2]=0.1-X

Ka=\frac{X*X}{0.1-X}

4.36*10^{-4}=\frac{X^2}{0.1-X}

4.36*10^{-5}-4.36*10^{-4}*X=X^2

Solving:

X=6.4*10^{-3}

The NaOH is in excess:

[NaOH]_f=0.2-0.0064=0.1936

pH=14+log([OH^-])

pH=14+log(0.1936)=13.28

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Mn2+
snow_lady [41]

The molarity of the stock Mn²⁺ ions is 0.0288 M

Based on the dilution formula;

  • The molarity of A is 0.00144 M
  • The molarity of B is 0.0000576 M
  • The molarity of C is 0.000001152 M

<h3>What is the molarity of a solution?</h3>

The molarity of a solution is the number of moles of a solute dissolved in a given volume of solution in liters.

  • Molarity = number of moles/volume

The molarity of the stock solution is:

moles of Mn²⁺ ions = mass / molar mass

molar mass of  Mn²⁺ ions = 55.0 g/mol

moles of Mn²⁺ ions = 1.584 / 55

moles of Mn²⁺ ions = 0.0288 moles

molarity of Mn²⁺ ions = 0.0288 / 1

molarity of Mn²⁺ ions = 0.0288 M

The dilution formula is used to determine the molarities of A, B, and C.

C₁V₁ = C₂V₂

C₂ = C₁V₁ / V₂

Where;

  • C₁ = initial molarity
  • V₁ = initial volume
  • C₂ = final molarity
  • V₂ = final volume

Molarity of A = 50 * 0.0288 / 1000

Molarity of A = 0.00144 M

Molarity of B = 10 * 0.00144 / 250

Molarity of B = 0.0000576 M

Molarity of C = 10 * 0.0000576 / 500

Molarity of C = 0.000001152 M

Learn more about molarity at: brainly.com/question/17138838

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