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 )

a. hope i'm right

explanation:

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.