THE LAST ONE!!

This problem set is due by $10$ am on Monday, Dec. $18,2023.$

This problem involves the roles of acetic acid and acetate ion in aqueous solutions.

a$.$ Consider a solution of sodium acetate. Write down the relevant hydrolysis reaction.

Identify the base, acid, conjugate acid, and conjugate base.

b$.$ Consider a solution of acetic acid. Write down the relevant hydrolysis reaction.

Identify the acid, base, conjugate base, and conjugate acid.

c$.$ Write down the equilibrium$-$constant expression $(K_b)$ for acetate. Just use symbols;

numbers are not required.

d$.$ Write down the equilibrium$-$constant expression $(K_a)$ for acetic acid. Just use

symbols; numbers are not required.

e$.$ Use your answer from part $d$ to write down an expression $($no numbers$)$ for $\frac{K_w}{K_a}.$

Show that this expression equals $K_b$ for acetate.

Methylamine is a weak base analogous to ammonia. It has a methyl group in place of

one of the hydrogens of ammonia. The $K_b$ value of methylamine is $4\mathrm{.}4{10}^{-4}.$ Methylam$-$

monium is the conjugate acid of methylamine. Determine the $pH$ of a $0\mathrm{.}20M$ solution of

methylammonium chloride.

Lactic acid is a carboxylic acid with a similar structure to acetic acid. You can abbreviate

lactic acid as HLac. Its conjugate base is lactate ion, abbreviated as $La{c}^{-}.$ The $K_a$ value

of lactic acid is $5\mathrm{.}6{10}^{-4}.$

a$.$ Suppose you make $1\mathrm{.}00L$ of a solution that contains both lactic acid and sodium

lactate, each with an initial concentration of $0\mathrm{.}25M.$ Calculate the $pH$ of this solution.

b$.$ Calculate the new $pH$ if you add $0\mathrm{.}020$mol of $HCl$ to the solution in part a$.$ Assume

the volume does not change.

c$.$ Starting with the original solution in part a$,$ suppose you add $0\mathrm{.}020$molNaOH.

Assume the volume does not change. Calculate the new $pH.$

d$.$ What is the $pH$ if you add $0\mathrm{.}020$mol of $HCl$ to $1\mathrm{.}00L$ of pure water? Assume the

volume does not change.

e$.$ What is the $pH$ if you add $0\mathrm{.}020$mol of NaOH to $1\mathrm{.}00L$ of pure water? Assume the

volume does not change.

f$.$ Explain in a sentence or two why the $pH$ changes in d$.$ and e$.$ are much more drastic

than the $pH$ changes in $b.$ and $c.$