could someone please help from 1 to 4 CHEM 132 Practice Problems 1. A 60.0 mL sa
ID: 697704 • Letter: C
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could someone please help from 1 to 4
CHEM 132 Practice Problems 1. A 60.0 mL sample of 0.20 M HF is added to a 0 10 M KOH. Determine the pH of the solution after the addition of 0.0012-)-437 100.0 mL of KOH. The Ka of HF is 3.5 10-4 t ) 2. Determine the pH of the resulting solution after the addition of 15 mL of 0.25 M HC to the solution in 1) 3. What is the pH of a solution prepared by adding 25.00 mL of 0.10 M CH3CO2H to 25.00 mL of 0 0 10 M CH3CO2Na? Ka = 1.8 x 10-5 for CH3CO-H 4. What is the pH of a solution prepared by adding 50.00 mL of 0.10 M methylamine, CH3NH2, to 20.00 mL of 0.10 M methylammonium chloride, CH3NH3CI? Kb 3.70x 10-4 for methylamine. Write an equilibrium reaction that takes place in this system.Explanation / Answer
1)
pH = pKa + log([salt]/[Acid])
pH = -log(3.5*10^-4)+ log((100*0.1)/(0.2*60))
pH = 3.377
2)
number of moles of HCL = 15 * 0.25 = 3.75 mmoles
after adding HCl pH of the solution is given by,
pH = pKa + log([salt]-[HCl]/[acid]+[HCl])
pH = -log(3.5*10^-4) + log(((100*0.1)-(3.75))/((60*0.2)+(3.75)))
pH = 3.0545
3)
pH = pKa + log (salt/Acid)
pH = -log(1.8*10^-5) + log((25*0.01)/(25*0.1))
pH = 3.7447
4)
pOH = pKb + log (salt/Base)
pOH = -log(3.7*10^-4) + log((20*0.1)/(50*0.1))
pOH = 3.0338
pH = 14 - 3.0338 = 10.9662
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