How many mol of KOH(s) must be added to 200.00 mL of pure water to make a solute

Question

How many mol of KOH(s) must be added to 200.00 mL of pure water to make a solute of pH 12.85? Assume no volume change upon addition of the KOH. A. 0.014 mol B. 0.029 mol C. 0.070 mol D. 0.145 mol E. 0.233 mol The solubility product K_sp for Pb(IO_3)_2 is equal to 2.6 times 10^-13 at 25 degree C. What is the molar solubility of Pb(IO_3)_2 in pure water at 25 degree C? A. 5.1 times 10^-7 mol/L B. 4.0 times 10^-5 mol/L C. 6.0 times 10^-7 mol/L D. 6.5 times 10^14 mol/L E. 5.1 times 10^-6 mol/L Given: Au^3+ + 3e^- rightarrow Au(s) E degree = 1.40 V Cu^2+ + 2e- rightarrow Cu(s) E degree = 0.34 V calculate Delta G degree for the reaction 3 Cu(s) + 2 Au^3+(aq) rightarrow 3 Cu^2+ (aq) + 2 Au(s) A. -205 kJ B. -614 kJ C. 205 kJ D. -307 kJ E. 614 kJ Which of following sparingly soluble compounds would not be more soluble in dilute nitric acid than in pure water? A. PbCO_3 B. Fe(OH)_2 c. AgI D. MgF_2 E. Ag_2S

Explanation / Answer

KOH K+ +OH-

PH = 12.85

POH =14-12.85 =1.15

[OH] = 10[-1.15] =0.070 M

[KOH] = 0.070 M

VOL = 200ML

No. of moles =molarity x volume (L)

= 0.070 M x 0.200L = 0.014 mol

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