Calculate the pH of the resulting solution if 34.0 mL of 0.340 M HCl(aq) is adde

Question

Calculate the pH of the resulting solution if 34.0 mL of 0.340 M HCl(aq) is added to

o 4/11/2016 1 1 :55 PM 30.8/504/8/2016 07:45 PM Gradeboolk Calculator CalculatorPeriodic Table -Periodic Table Question 16 of 20 Incorrect Map tion Donald McQuarrie Peter A. Rock Ethan Gallogly University Science Books presented by Sapling Learning eneral Chemistr Calculate the pH of the resulting solution if 34.0 mL of 0.340 M HCl(aq) is added to (a) 39.0 mL of 0.340 M NaOH(aq) Number | 11.24 pH= (b) 44.0 mL of 0.390 M NaOH(aq) Number pH= 11 11.75 There is a hint available View the hint by clicking on the bottom divider bar. Click on the divider bar again to hide the hint. Close PreviousS Exit Hint

Explanation / Answer

its a strong acid vs strong base

NaOH + HCl ----> NaCl + H2O

one eq of NaOH will react with one eq of HCl

no of moles of HCl = 0.34 M x 0.034 L = 0.01156 mol

part a

no of moles of NaOH 0.34 M x 0.039 L = 0.01326 mol

0.01156 moles of HCl will react with 0.01156 mol of NaOH

moles of NaOH remaining = 0.01326 - 0.01156 = 0.0017 mol of NaOH

total volume = 39 + 34 = 73 mL = 0.073 L

concentration of NaOH remaining = 0.0017 mol / 0.073 L = 0.02328 M

pOH = -log(0.02328) = 1.63

pH = 14-pOH = 14-1.63 = 12.36

Part B

moles of NaOH = 0.39 M x 0.044 L = 0.01716 mol

0.01156 moles of HCl will react with 0.01156 mol of NaOH

moles of NaOH remaining = 0.01716 - 0.01156 = 0.0056 mol

total volume = 34 + 44 = 78 mL = 0.078 L

concentration of NaOH remaining = 0.0056 mol / 0.078 L = 0.0718 M

pOH = -log(0.0718) = 1.144

pH = 14-pOH = 14-1.14 = 12.85

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