CHEM 1310-Spring16-MIGYANK Activities and Due Dates HW 19a /2016 12:00 AM 81/100

Question

CHEM 1310-Spring16-MIGYANK Activities and Due Dates HW 19a /2016 12:00 AM 81/100 /2016 12:00 AM 4 81/100 Print Calculator -Print n 15 of 21 11 Calculator -a Periodic Table Periodic Table University Science Books presented by Sapling Leamin ld McQuarrie Peter A. Rock Ethan Gallogly Diculate the pH of the resulting solution if 31.0 mL of 0.310 M HC(aq) is added to (a) 41.0 mL of 0.310 M NaOH(aq) Number (b) 21.0 mL of 0.410 M NaOH(aq). Number Hint O Previous Gve Up & View Solution Check Answer Speed up browsing by disabling add-ons

Explanation / Answer

a) NaOH + HCl ---- NaCl + H2O

Number of moles of HCl = 31/1000 * 0.310 = 0.00961 moles

Number of moles of NaOH = 41/1000 * 0.310 = 0.01271

Excess moles of NaOH = 0.01271 - 0.00961 = 0.0031 moles

Molarity of [OH-] in the solution = number of moles/volume of solution = 0.0031/(62/1000) = 0.05M

pOH = -log[OH-] = --log[0.05] = 1.3010

pH = 14 - 1.3010 = 12.69

b) NaOH + HCl ---- NaCl + H2O

Number of moles of HCl = 31/1000 * 0.310 = 0.00961 moles

Number of moles of NaOH = 21/1000 * 0.410 = 0.0861 moles

Excess moles of HCl = 0.00961 - 0.00861 = 0.001 moles

Molarity of [H+] in the solution = number of moles/volume of solution = 0.001/(52/1000) = 0.01923M

pH = -log[H+] = --log[0.01923] = 1.716

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