How much 10.0 m hno3 must be added to 1.00 l of a buffer that is 0.0100 m acetic acid and 0.100 m sodium acetate to reduce the ph to 5.05 ?

Initial moles of CH₃COOH = 1.00 x 0.0100 = 0.0100 mol

Initial moles of CH₃COONa = 1.00 x 0.100 = 0.100 mol

Let a be the volume of HNO₃ that must be added

Moles of HNO₃ added = a/1000 x 10.0 = 0.01a mol

CH₃COONa + HNO₃ => CH₃COOH + NaNO₃

Moles of CH₃COOH = 0.0100 + 0.01a

Moles of CH₃COONa = 0.100 - 0.01a

pH = pKa + log([CH₃COONa]/[CH₃COOH])

= pKa + log(moles of CH₃COONa/moles of CH₃COOH)

5.05 = 4.75 + log((0.100 - 0.01a)/(0.0100 + 0.01a))

log((0.100 - 0.01a)/(0.0100 + 0.01a)) = 0.3

(0.100 - 0.01a)/(0.0100 + 0.01a) = 10^0.3 = 2.0

0.03 a = 0.08

a = 2.67

Volume of HNO3 = a = 2.13 mL

Elias is almost correct. he just got it backwards.the correct answer is that it demonstrates elastic potential energy transforming into kinetic energy.

257 grams h2o i think