What is the final pH when 42.1 mL of 3.2 M NaOH is added to 1.00 L of 0.150 M HF

Question

What is the final pH when 42.1 mL of 3.2 M NaOH is added to 1.00 L of 0.150 M HF (K_a = 3.5 times 10^-4) Why is the molar solubility of silver chloride lower in an aqueous solution of sodium chloride than in pure water? because silver ions and sodium ions form a completely insoluble precipitate and fall out of the solution because sodium chloride takes up all the water, and doesn't leave any interactions for the silver chloride, causing reduced solubility because the presence of extra chloride shifts the equilibrium of the AgCl dissolution reaction toward the undissociated (solid) silver chloride What is the pH of a buffer that is 0.058 M HNO_2 and 0.032 M NaNO_2? The K_a for HNO_2 is 4.0 times 10^-4. The strong acid HNO_3 is titrated with the strong base KOH. What substance(s) are present at the equivalence point?

Explanation / Answer

Q13.

final pH for:

mol of HF = MV = 0.15*1 = 0.15 mol

mol of NaOH = MV = 3.2*(42.1*10^-3) = 0.13472 mol

so, after reaction

acid left = 0.15-0.13472 = 0.01528 mol

mol of conjguate = 0.13472 mol

pH = pKa + log(F-/HF)

pKa = -log(3.5*10^-4) = 3.45593

pH = 3.45593+ log(0.13472 /0.01528 ) = 4.40123

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