I really neeed help

when 1.00 g of stearic acid ( a saturated fatty acid) is dissolved in water, then diluted to exctly 10.00 ml, the PH of the resulting solution is 2.73.the fornula weight of steric acid is 284.5. calculate ka and pka for this acid

Ka=

Pka=

First calculate the molarity of the solution of acid:
1.0g in 10.0mL = 100g in 1000mL
Molar mass = 284.5
mol of stearic acid = 100/284.5 = 0.351 moles in 1000 mL
Molarity = 0.351M

Calculate [H+]
pH = 2.73
[H+] = 10^-pH
[H+] = 10^-2.73
[H+] = 1.86*10^-3

Use the Ka equation to calculate Ka - call stearic acid HA
Ka = [H+] [A-] / [HA]
Because [H+] = [A-] we can write [H+][A-] = [H+]²
Because [HA]>>>> [H+] we take [HA] = 0.351
Substitute
Ka = [H+]² / 0.351
Ka = (1.86*10^-3)²/0.351
Ka = 3.467*10^-6 / 0.351
Ka = 9.88*10^-6

pKa = - log Ka = -log (9.88*10^-6) = 5.0

Note: are you sure about this question - it deals with a stearic acid solution of 100g/L Stearic acid is almost insoluble in water ≈ 3mg / L at 20°C. Because of this very low solubility I do not think that Ka or pKa values are relevant - I could not find any published values.

you use formula pH=(pka-log c)/2 or pka =2pH+log c

c is molarity of the acid if monovalent
you have 1g in 10mL , so in a liter 1000mL you have 1*1000/10=100g
the molarity is 100/284.5=0.352M log 0.352= -0.454

pKa= 5.46- (-0.456)=5.91

ka= 10^(-5.91)=1.22*10^-6