For the ionization of a weak acid, HA:

a) give the expression for Ka. b) after taking the log 10 of both sides ofthe quation in "a" sp;ve fpr pH. c) under what conditions would pH be equal to pK?

This is just the derivation of the Henderson-Hasselbalch equation.

I'll get you started.
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Now do what b says. Take the log of both sides.
log Ka = log(H^+)(A^-)/(HA) or
log Ka = log[(A^-)/(HA)] + log(H^+)
Multiply through by -1
-log Ka = - log([A^-)/(HA)] -log(H^+)
But note that -log Ka = pKa and -log(H^+) = pH. Now you finish. You should end up with
pH = pKa + log[(A^-)/(HA)]
Post your work if you get stuck.
Obviously, pH = pKa when [(A^-)/(HA)] = 1

thank you!

a) The expression for Ka (acid dissociation constant) for the ionization of a weak acid HA can be given as:

Ka = [H+][A-] / [HA]

b) After taking the logarithm (log10) of both sides of the equation for Ka, we can solve for pH:

log10(Ka) = log10([H+][A-] / [HA])

pKa = -log10(Ka) = -log10([H+][A-] / [HA])

pH = -log10([H+])

c) pH will be equal to pKa when the concentration of the weak acid and its conjugate base are equal, meaning [HA] = [A-]. This occurs at the half-equivalence point during a titration, where half of the weak acid has been converted into its conjugate base.

a) The ionization of a weak acid, HA, can be represented by the chemical equation: HA ⇌ H+ + A-. The expression for the acid dissociation constant, Ka, is given by the ratio of the concentrations of the products (H+ and A-) to the concentration of the undissociated acid (HA). Therefore, the expression for Ka is:

Ka = [H+][A-] / [HA]

b) To find the pH after taking the logarithm base 10 (log10) of both sides of the equation in part "a," you apply the negative logarithm of the concentration of H+ ions on each side. This yields the equation:

-log10(Ka) = -log10([H+][A-] / [HA])

Since the logarithm of the product is equal to the sum of the logarithms of the individual terms, the equation simplifies to:

-pKa = -log10([H+]) - log10([A-] / [HA])

The negative logarithm of [H+] is equal to the pH of the solution. Therefore, rearranging the equation, we get:

pH = pKa - log10([A-] / [HA])

c) The pH is equal to pK (the negative logarithm of the acid dissociation constant) under specific conditions. When the concentration of the acid (HA) is equal to the concentration of its conjugate base (A-), the log10([A-] / [HA]) term in the equation from part "b" becomes zero. Consequently, the pH equals pK. This occurs at the midpoint of acid-base titrations when a weak acid is halfway neutralized by a strong base or when a weak base is halfway neutralized by a strong acid.