Question

Calculate the hydronium ion concentration and the pH of the solution that results when 28.5 mL of 0.16 M acetic acid, HCH3CO2H (Ka= 1.8x10-5), is mixed with 1.3 mL of 0.26 M NaOH

Hydronium ion concentration = _______ M

pH = _____

****PLEASE ONLY ATTEMPT IF YOU KNOW THE ANSWER AS I'M DOWN TO MY LAST ATTEMPT*** THANKS!

Answer 1

milimoles of CH3COOH = 4.56 [ by formula Molarity =no of moles / volume ]

milimoles of NaOH = 0.338 [ since volume is in mL milimoles will come if volume would be in L

pKa = -logKa then only moles will come]

= -log(1.8*10^-5)

= 4.74

[Acetic acid is weak acid and NaOH os strong base so NaOH will dissociate completely and Acetic acid will dissociate partially]

  

Initial conc 4.56 0.338 0 0

After salt formation 4.22 0 0.338 0

pH = pKa +log[salt] / [acid]

= 4.74 + log 0.338 / 4.22

= 4.74 -1.096

= 3.644

[H₃O^{+] =} 10^-pH

⁼ 10^-3.64

   ⁼ 2.29 * 10 ^-4 M