If you add 5.00 ml of 0.100 m sodium hydroxide to 50.0 ml of acetate buffer that is 0.100 m in both acetic acid and sodium acetate, what is the ph of the resulting solution? acetic acid: ka = 1.8. x 10-5

see explanation ...

Explanation:

5.00ml(0.100M NaOH) + 0.100M HOAc/NaOAc Bfr

5.00ml(0.100M NaOH) = 0.005(0.100) mole NaOH = 0.0005 mole NaOH = 0.0005 mole OH⁻ in 55 ml Bfr solution (50ml + 5ml)=> [OH⁻] = (0.0005/0.055)M OH⁻ = 0.0091M OH⁻ ≈ 0.010M OH⁻ added into Bfr solution. The amount of OH⁻ added must be removed by H⁺ in the HOAc equilibrium; that is,  H⁺ + OH⁻ → H₂O leaving a void at the H⁺ position in the HOAc equilibrium. HOAc then decomposes to replace the H⁺ removed by the excess OH⁻ giving new H⁺ and OAc⁻ concentrations. Reaction shifts right => subtract 0.01M from HOAc side of equilibrium and add 0.01M to OAc⁻side of equilibrium and recompute the H⁺ concentration and new pH.

                 HOAc  ⇄        H⁺     +    OAc⁻

C(i)             0.10M         ~0M*         0.10M =>pH=-log(Ka)=-log(1.85x10⁻⁵)=4.73

ΔC              -0.01M          +x          +0.01M

C(eq)          0.09M            x             0.11M  => New HOAc equil. conc.

Ka = [H⁺][OAc⁻]/[HOAc]

=> x(0.11)/(0.09) = 1.85x10⁻⁵

=> x = [H⁺] = 0.09(1.85x10⁻⁵)/0.11 = 1.52x10⁻⁵M (after adding NaOH)

=> pH = -log[H⁺] -log(1.52x10⁻⁵) = 4.82 ( pH shifts to more basic value b/c of OH⁻ addition )

i think it is d. e because the next one is igneous.

answer; i believe the correct answer is (solid gold);