Question

How much 10.0 M HNO 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 4.95? 3.87 mL 2nd attempt 1st attempt C TRY AG VIEW SOLUTION 13 OF 22 QUESTIONS COMPLETED SAMSUNG

Answer 1

Step-by-step explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

Moles of acetic acid = 0.0100 M × 1.00 L

= 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

= 0.100 mol

When
`HNO_(3)` is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles
`HNO_(3)`)

= 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles
`HNO_(3)`)

= 0.100 mol - x

According to Henderson - Hasselbalch equation,

pH =
`pK_(a) + log([conjugate base])/([weak acid])`

pH =
`pKa + log((new moles of sodium acetate))/((new moles of acetic acid))</p><p>`

4.95 = 4.75 +
`log((0.100 mol - x))/((0.0100 mol + x))`

`log((0.100 mol - x))/((0.0100 mol + x))` = 4.95 - 4.75

= 0.20

`((0.100 mol - x))/((0.0100 mol + x))` = antilog (0.20)

= 1.6

Hence, x = 0.032555 mol

Therefore, moles of
`HNO_(3)` = 0.032555 mol

volume of
`HNO_(3)` =
`(moles HNO_(3))/(molarity of HNO_(3))`

=
`(0.032555 mol)/(10.0 M)`

= 0.0032555 L

or, = 3.25 (as 1 L = 1000 mL)

Thus, we can conclude that volume of
`HNO_(3)` added is 3.26 mL.