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nevsk [136]
3 years ago
11

Calculate the pOH and the pH of a 5.0 x 10-2 M solution of NaOH.

Chemistry
1 answer:
k0ka [10]3 years ago
5 0

Answer:

pOH = 1.3, pH = 12.7

Explanation:

Since NaOH is a strong base, it will completely ionize; further, since it completely ionizes, our hydroxide concentration (a product of the ionization) will be the same as the given concentration of NaOH.

NaOH -> Na⁺ + OH⁻, [OH⁻] = 5.0 x 10^-2 M

pOH is the negative log of the hydroxide concentration, so plug our hydroxide concentration in:

pOH = -log[OH⁻] = -log[5.0 x 10^-2 M] = 1.3

Since pH + pOH = 14, we can plug in pOH and solve for pH:

pH + 1.3 = 14

pH = 14 - 1.3 = 12.7

Thus, our pOH = 1.3 and pH = 12.7.

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H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O

Formula of molarity is:

Molarity = \frac{Moles of solute}{Volume of solution in Liters}

Molarity = 6.0 M, Volume = 35 mL = 0.035 L

Substituting the values,

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Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole

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Molar mass of NaHCO_3 = 84.007 g/mol

So, the mass of NaHCO_3 = 84.007 g/mol \times 0.42 mol = 35.283 g




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