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2. In a second titration with the same solution of NaOH as used in Question #1, the student weighs out a sample of KHP of 0.359 g. Calculate the volume of the NaOH solution needed to neutralize this sample of KHP. moles of KHP = 0.359 g / 204 g/mol = 1.76 X 10^-3 mole vol. of NaOH = 1.76 X 10^-3 / 0.0873 = 0.0202 L = 20.2 mL 3. A monoprotic weak acid with the general formula of HA will react with a base, such as NaOH. Write the neutralization equation which describes the reaction. HA(aq) + NaOH(aq) <=> NaA(aq) + H2O(l) or HA(aq) + Na+(aq) + OH-(aq) <=> Na+(aq) + A-(aq) + H2O(l) 4. If Ka for the weak acid, HA is 1.8 x 10^-5, calculate the magnitude of the equilibrium constant for the equation in Question #3. (Note: First write the net ionic equation for the equation in Question #3.) 1) Keq = [A-][H3O+] / [HA][OH-][H3O+] From 1 and 2 Keq = 10^-14 so, Keq = 1.8 X 10^9 2) Ka = [A-][H3O+] / [HA] = 1.8 X 10^-5 5. Another Chem 1515 student was given a 50.0 mL sample of the weak acid in Problem #4. The concentration of the weak acid is given as 0.15 M. Using a pipette, she delivers 25.0 mL of the weak acid into a 100 mL beaker. She then titrated this solution using a standardized solution of 0.21 M NaOH. a) How many moles of the weak acid were added to the beaker? moles of weak acid = 0.15 / (25 / 1000) = 3.75 X 10^-3 moles b) How many moles of NaOH are required to neutralize (reach the equivalence point) the sample of weak acid? Since, 1 mol of HA = 1 mol NaOH so, moles of NaOH required = 3.75 X 10^-3 moles.

          2. In a second titration with the same solution of NaOH as used in Question #1, the student weighs out a sample of KHP of 0.359 g. Calculate the volume of the NaOH solution needed to neutralize this sample of KHP. moles of KHP = 0.359 g / 204 g/mol = 1.76 X 10^-3 mole vol. of NaOH = 1.76 X 10^-3 / 0.0873 = 0.0202 L = 20.2 mL 3. A monoprotic weak acid with the general formula of HA will react with a base, such as NaOH. Write the neutralization equation which describes the reaction. HA(aq) + NaOH(aq) <=> NaA(aq) + H2O(l) or HA(aq) + Na+(aq) + OH-(aq) <=> Na+(aq) + A-(aq) + H2O(l) 4. If Ka for the weak acid, HA is 1.8 x 10^-5, calculate the magnitude of the equilibrium constant for the equation in Question #3. (Note: First write the net ionic equation for the equation in Question #3.) 1) Keq = [A-][H3O+] / [HA][OH-][H3O+] From 1 and 2 Keq = 10^-14 so, Keq = 1.8 X 10^9 2) Ka = [A-][H3O+] / [HA] = 1.8 X 10^-5 5. Another Chem 1515 student was given a 50.0 mL sample of the weak acid in Problem #4. The concentration of the weak acid is given as 0.15 M. Using a pipette, she delivers 25.0 mL of the weak acid into a 100 mL beaker. She then titrated this solution using a standardized solution of 0.21 M NaOH. a) How many moles of the weak acid were added to the beaker? moles of weak acid = 0.15 / (25 / 1000) = 3.75 X 10^-3 moles b) How many moles of NaOH are required to neutralize (reach the equivalence point) the sample of weak acid? Since, 1 mol of HA = 1 mol NaOH so, moles of NaOH required = 3.75 X 10^-3 moles.

2. In a second titration with the same solution of NaOH as used in Question #1, the student weighs out a sample of KHP of 0.359 g. Calculate the volume of the NaOH solution needed to neutralize this sample of KHP. moles of KHP = 0.359 g / 204 g/mol = 1.76 X 10^-3 mole vol. of NaOH = 1.76 X 10^-3 / 0.0873 = 0.0202 L = 20.2 mL 3. A monoprotic weak acid with the general formula of HA will react with a base, such as NaOH. Write the neutralization equation which describes the reaction. HA(aq) + NaOH(aq) <=> NaA(aq) + H2O(l) or HA(aq) + Na+(aq) + OH-(aq) <=> Na+(aq) + A-(aq) + H2O(l) 4. If Ka for the weak acid, HA is 1.8 x 10^-5, calculate the magnitude of the equilibrium constant for the equation in Question #3. (Note: First write the net ionic equation for the equation in Question #3.) 1) Keq = [A-][H3O+] / [HA][OH-][H3O+] From 1 and 2 Keq = 10^-14 so, Keq = 1.8 X 10^9 2) Ka = [A-][H3O+] / [HA] = 1.8 X 10^-5 5. Another Chem 1515 student was given a 50.0 mL sample of the weak acid in Problem #4. The concentration of the weak acid is given as 0.15 M. Using a pipette, she delivers 25.0 mL of the weak acid into a 100 mL beaker. She then titrated this solution using a standardized solution of 0.21 M NaOH. a) How many moles of the weak acid were added to the beaker? moles of weak acid = 0.15 / (25 / 1000) = 3.75 X 10^-3 moles b) How many moles of NaOH are required to neutralize (reach the equivalence point) the sample of weak acid? Since, 1 mol of HA = 1 mol NaOH so, moles of NaOH required = 3.75 X 10^-3 moles.

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