Please answer the following question. For problems requiring calculations, be su
ID: 1041149 • Letter: P
Question
Please answer the following question. For problems requiring calculations, be sure to SHOW ALL WORK and GIVE ANSWERS WITH THE CORRECT NUMBER OF SIGNIFICANT FIGURES AND UNITS. Full credit will not be given for a correct answer unless all of your work is shown. POINTS WILL BE SUBTRACTED if an answer does not have the correct number of significant figures and units (5 points). Calculate the pH of a solution prepared by mixing 3.00 grams of butyric acid (HC H:02) with 0.75 grams of NaOH in water. 02) with 0.75 grams of NaOH in water. The Ka of butyric acid is 1.5 x 10.Explanation / Answer
mass of Butyric acid= 3.00 grams
molar mass of butyric acid=88.11 gram/mole
number of moles of Butyric acid= mass/molar mass = 3.00/88.11= 0.0340 moles
mass of NaOH= 0.75 grams
molar mass of NaOH= 40.0gram/mole
number of moles of NaOH= 0.75/40,0= 0.01875 moles
CH3CH2CH2COOH + NaOH ------------------- CH3CH2CH2COONa + H2O
0.0340 0.01875 0
-0.01875 -0.01875 +0.01875
0.01525 0 +0.01875
Ka= 1.5x10^-5
-log(Ka)= -log(1.5x10^-5)
PKa= 4,82
PH= PKa + log[salt]/[acid]
PH= 4.82 + log(0.01875/0.01525)
PH= 4.909
PH= 4.91