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A solution is prepared by adding 0.0150 moles of NaX to 50.0 mL of 0.100 M...

A solution is prepared by adding 0.0150 moles of NaX to 50.0 mL of 0.100 M HCl. This solution, which still occupies 50.0 mL, has a pH of 4.809.

A) Calculate the concentrations of H+, X-, and HX.

B) Determine the ionization constant for the acid HX.

C) Calculate the pH of this solutiona fter 10.0 mL of 0.500 M NaOH has been added. Assume the volumes are additive.

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Answer #1

NaX + HCl NaCl + HX

No. of moles of HCl = M V = 0.100 0.05 = 0.005 moles

So HCl is the limiting reactant

pH = 4.809 = -log[H+]

[H+] = 1.55 10-5 M

1 mole of HCl gives 1 mole of HX

Hence, 0.005 moles of HCl will give 0.005 moles of HX

[HX] = n/V = 0.005/0.05 = 0.1 M

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