AP acid base equilibria questions?

you can convert the concentrations to moles and make calculations easier hasselbachs equation works with either.

your acid is HC2H3O2 and you have 4.8E-3 moles of this

your base is NaC2H3O2 and you have 9E-3 moles of this

Now you use the hasselbach equation which states that

pH=Pka+log([base]/[acid])

therefor plugging in your known values you get

pH=-log(1.8E-5)+log(9E-3/4.8E-3)

pH=5.01

addition of a strong base with a common ion will shift the equation to the products side thus you will use up all your NaOH so take you moles of NaOH which is equal to 1E-3 mole and subtract it from your acid component HC2H3O2 and add it to your base component NaC2H3O2

HC2H3O2=(4.8E-3)-(1E-3)=3.8E-3 mole

NaC2H3O2=(9E-3)+(1E-3)=1E-2 mole

now plug your new moles into the equation to get

pH=-log(1.8E-5)+log(1E-2/3.8E-3)

pH=5.16

be sure to use correct sig figs and dont round until the end

Good luck on your AP exams hope you do well

sincerly yours

--Former AP student current college Chem major

i choose to preface my answer by potential of telling u that u can e mail me with the different questions. it isn't basic to respond to all 3 questions, yet ill attempt. a million. we see that H3O+ is the most valuable acid, because water is a very very susceptible base. thye conjugate of a good acid is a susceptible base. the superior the acid, the weaker its conjugate. the superior the bottom, the weaker its conjugate. 2. i recognize for in basic terms about positive that B(OH)3 is an ACID. usually, all nonmetal OH's are acids. if u rearrange this formula, u get H3BO3. i am going to say D. 3. to discover the pH of a susceptible acid, u take the sq. root of Ka x Molarity. considering the fact that all molarities are a million, take the sq. root of all the Ka's and take the adverse log of what u were given. and the nearest one to 7 is the winner. it might want to correctly be the NH4+ i wish this helps.