A 35. mL of a 0.045 M solution of acetic acid, pKa = 4.74, is titrated...

A 35. mL of a 0.045 M solution of acetic acid, pKa = 4.74, is titrated with a 0.068 M solution of NaOH. What is the hydroxide ion concentration of the resulting solution after 30.0 mL of NaOH is added?

Expert Solution

no of moles of CH3COOH = molarity * volume in L

= 0.045*0.035 = 0.001575 moles

no of moles of NaOH = molarity * volume in L

= 0.068*0.03 = 0.00204 moles

no of moles of excess NaOH = 0.00204-0.001575 = 0.000465 moles

total volume = 35+30 = 65ml = 0.065L

molarity of NaOH excess = no of moles/total volume

= 0.000465/0.065 = 0.00715M

NaOH -----------------> Na^+ + OH^-

0.00715M 0.00715M

[OH^-] = 0.00715 M >>>>answer

queen_honey_blossom answered 2 years ago

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