Asked by Teagan
Calculate the volume of 0.05 M NaOH which must be added to 100 cm3 of 0.05 M methanoic acid to obtain a buffer of pH 4.23. (Ka for methanoic acid = 1.77 x 10-4)
Answers
Answered by
DrBob222
Use the Henderson-Hasselbalch equation.
millimols HCOOH = mL x M = 100 x 0.05 = 5
......HCOOH + NaOH ==> HCOONa + H2O
I.......5.......0........0........0
add.............x................
C.......-x.....-x........+x
E.......5-x......0........x
x = base
5-x = acid
Substitute into the HH equation.
pH = pKa + log (base)/(acid)
4.23 = pKa + log (x)/(5-x)
Solve for x which is millimols NaOH that must be added. Then millimols NaOH = mL x M. You have mmols and M, solve for mL.
millimols HCOOH = mL x M = 100 x 0.05 = 5
......HCOOH + NaOH ==> HCOONa + H2O
I.......5.......0........0........0
add.............x................
C.......-x.....-x........+x
E.......5-x......0........x
x = base
5-x = acid
Substitute into the HH equation.
pH = pKa + log (base)/(acid)
4.23 = pKa + log (x)/(5-x)
Solve for x which is millimols NaOH that must be added. Then millimols NaOH = mL x M. You have mmols and M, solve for mL.
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