Chem. 111 Show all your set-ups. Significant figures must be correct. Ka of Weak

Question

Chem. 111 Show all your set-ups. Significant figures must be correct. Ka of Weak Acid Name: ID) A 30.00 mls solution of 0.1500 M Sodium acetate is added 10.0 mls of 0.250 M Acetic acid The measured pH of the resulting solution is 4.33, Calculate the Ka of Acetic acid Hint: Calculate the new molar concentrations of each component in the solstion, the equilibrium equation NITYAL CHANGE (10 pts.) 2. A 10.00 g solid salt of Nax (molar mass = 89.54 ghnol) is added and dissolved into a 5000 mL of a 2.00 M HHX solution (Ka of acid HX is 2.75 x 10-7) HX(aq) +H2O (I) Hjot (aq) + X. (aq) NITIA CHLANGS a. Calculate the molar concentration of [H,0] b. Calculate the pH c. Calculate the molar concentration of (OH-]

Explanation / Answer

Moles = Molarity* Volume in L, 1000ml= 1L

Moles of acetic acid = 0.25*10ml*1L/1000ml= 0.0025

Moles of sodium acetate= .15*30ml*1L/1000ml= 0.0045

Volume of resulting solution = 30+10=40ml, 40/1000L

Concentrations : acetic acid =0.0025*1000/40, sodium acetate= 0.0045*40/1000

Since pH= pKa+ log [sodium acetate/acetic acid]

4.33= pKa+ log [0.0045/0.0025)

4.33= pKa+0.255

pKa= 4.33-0.255= 4.075

Ka= 8.4*10-5

2. molea of NaX added = mass/molar mass = 10/89.54= 0.1116, moles of HX= 2*500/1000= 1

after mixing : concentrations : NaX= 0.1116/0.5 =0.2232 M and HX= 2M

pH= pKa+ log [NaX]/[HX], Ka= 2.75*10-7, pKa= 6.6

pH= 6.6+ log [0.2232/2)=5.65

pH= -log [H3O+]

[H3O+]= 10(-5.65) =2.25*10-6

pOH= 14-pH= 14-5.65= 8.35

[OH-]= 10(-8.35)= 4.45*10-9

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