Question

What is the pH of a solution that results from mixing together equal volumes of a 0.130 M solution of acetic acid and a 0.0650 M solution of sodium hydroxide

Answer 1

CH3-COOH -----------> CH3-COO- + H+

0.13o - - I

-x +x +x C

(0.003-x) +x +x E

Ka = 1.8 * 10^-5 = x²

  (0.130-x)

Ignoring x in the denominator as it is soto small [x represnets moles dissociated from a weaka cid]

x = 2.32 * 10^-6

= 0.00000232

represents moles of [H+] from acetic acid

moles of NaOH = 0.0650

These 0.0650 moles of NaOH will consume 0.00000232 of H+s

mole sof NaOH remaining = 0.0649

pOH = - log ( 0.0649)

pOH = 1.18

pH = 14 - 1.18 = 12.8

Answer 2

CH3-COOH -----------> CH3-COO- + H+

0.13o - - I

-x +x +x C

(0.003-x) +x +x E

Ka = 1.8 * 10^-5 = x²

  (0.130-x)

Ignoring x in the denominator as it is soto small [x represnets moles dissociated from a weaka cid]

x = 2.32 * 10^-6

= 0.00000232

represents moles of [H+] from acetic acid

moles of NaOH = 0.0650

These 0.0650 moles of NaOH will consume 0.00000232 of H+s

mole sof NaOH remaining = 0.0649

pOH = - log ( 0.0649)

pOH = 1.18

pH = 14 - 1.18 = 12.8