NaCl is added slowly to a solution din is 0.010 M each in aqueous Cu^+, Ag^+, an

Question

NaCl is added slowly to a solution din is 0.010 M each in aqueous Cu^+, Ag^+, and Au^+. The k_sp's for CuCI, AgCI, and AuCI are 1.9 times 10^-7, 1.8 times 10^10, and 2.0 times 10^-13, respectively. Which compound will precipitate first? A solution of 30.00 mL of 0.1250 M HCl(aq) is titrated with 0.1750 M NaOH(aq). A. Calculate the initial pH of the HCl solution. B. What volume of NaOH(aq) will be required to reach the equivalence point? C. What is the pH of the solution at the equivalence point? D. What is the pH after addition of 15.50 mL of NaOH solution?

Explanation / Answer

(D)

Moles of NaOH = 0.0155 L x 0.1750 M

=0.002713 moles

Moles of HCl = 0.030 L x 0.1250 M

=0.00375 moles

1 mole of NaOH reacts with 1 mole of HCl

So,

Moles of HCl left 0.00375 moles - 0.002713 moles == 0.0010375 moles

Volume = 15.5 mL + 30 mL = 0.0455 L

Concentration of H+ =  0.0010375 moles / 0.0455 L =0.02280 M

pH = -log (H+) = -log 0.02280

= 1.642

(E)

Moles of NaOH = 0.0375 L x 0.1750 M

=0.00656 moles

Moles of HCl = 0.030 L x 0.1250 M

=0.00375 moles

1 mole of NaOH reacts with 1 mole of HCl

So,

Here NaOH is in greater quantity than HCl

Moles of NaOH left 0.00656 moles - 0.00375 moles == 0.0028125 moles

Volume = 35.70 mL + 30 mL = 0.0657 L

Concentration of OH- =  0.0010375 moles / 0.0657 L =0.04281 M

pOH = -log (OH-) = -log 0.04281

= 1.368

pH = 14-pOH = 14 - 1.368 == 12.63

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