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

The density of water is 1 g/cm cubed and in order for the pebble to sink its density has to be greater than the density of water and in order for the twig to float its density has to be less than the density of water

Ithink is a because if say increased in electronegativity