# Calculate the pH after 0.020 mol NaOH is added to 1.00 L of each of the solutions below. [Solved]

certainly propanoate is the conjugate base of propanoic acid and can desire to be envisioned to hydrolyze and subsequently boost the pH.besides the incontrovertible fact that the presence of intense concentrations of OH-a million will supress the hydrolysis by employing shifting the equilibrium lower back in direction of propanoate propanoate-a million H2O ______ Propanoic-H OH-a million including extra OH-a million drives the above to the left by employing LeChateliers thought so calculation is as till now given

(a) HONH2 = weak base
(.020 x)(x)/(.100-x) = 1.1x 10^-8
OH- = .020.. pOH = 1.7
pH = 12.3

(d) pH = 6.18

HONH3 OH- > HONH3 H2O
I .121M .020M 0
C -.020M -.020M .020M
E .101M 0 .020M

pH = pKa log (.020/.101)
pKa = log Ka so Ka =1.0 x 10^-14 / 1.1 x 10^-8 = 9.09 x 10^-7
pKa = 6.04
pH = 6.04 (-.703) = 5.34

(c) pH= 12.3
reasoning OH- is strong base in water

certainly propanoate is the conjugate base of propanoic acid and can desire to be envisioned to hydrolyze and subsequently boost the pH.besides the incontrovertible fact that the presence of intense concentrations of OH-a million will supress the hydrolysis by employing shifting the equilibrium lower back in direction of propanoate propanoate-a million H2O ______ Propanoic-H OH-a million including extra OH-a million drives the above to the left by employing LeChateliers thought so calculation is as till now given

(d) pH = 6.18

This post is last updated on hrtanswers.com at Date : 1st of September – 2022