.1M of NaOH, I guessed that my mystery acid was potassium hydrogen. I used .2 gr

Question

.1M of NaOH,

I guessed that my mystery acid was potassium hydrogen. I used .2 grams of that.

My calculated ka was 3.16 x10-11

Molar mass is 210.53

The pH = pKa which is 4.75

9.5 Ml of NaOH was the volume of my equivalence point.

Thank you! If you need more inform let me know

i need help looking for the initial concentration of my acid and calculate the initial PH.

10. Calculate the initial concentration of your acid. Recall that the original volume of the solution was o.100 L [Acid]initial Using your calculated value of Ka and the initial concentration of the acid, calculate the initial pH of your acid. set up an ICE table. You may need to use the quadratic for

Explanation / Answer

Answer:

Hi student,

question is simple but confusing. any way i will explain in details.

weight of potassium hydrogen phthalate =0.2 g

molar mass=210.53

First calculate moles

moles= weight / Molar mass= 0.2/210.53=0.000949

volume of solution is given=0.100L

Calculate molarity

Molarity = Moles/Volume=0.000949/0.100=0.00949

Then write equation for change in concentration

KH -------> K+ + H-

Initial concentration 0.00949M 0 0

change in concentration -X M X M X M

equilibrium concentration 0.00949-X X M X M

Ka=[Product]/[Reactant]

Ka=[K+][H-]/[KH]

Ka=3.16 x10-11=[X][X]/[0.00949-X]

3.16 x10-11= X2/[0.00949-X]

X2 +3.16 x10-11 X - 3.00 X 10-13=0

Solve for x using  quadratic formula

X =5.44 X 10-7

Initial concentration=5.44 X 10-7

pH=-log[5.44 X 10-7]

pH=6.26

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