!! LONG ANSWER !!
The initial pH is "2.85"2.85 and the equivalence point pH is "8.87"8.87.
So, you start with a solution of acetic acid. The pH of this solution can be determined by using the concentration of protons in solution. For that, you'll have to set up an ICE chart
....CH_3COOH_((aq)) rightleftharpoons CH_3COO_((aq))^(-) + H_((aq))^(+)CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)
I.....0.1120..................................0.........................0
C....(-x).....................................(+x)......................(+x)
E...(0.1120-x)............................x...........................x
Use the expression of the acid dissociation constant to solve for xx
K_a = (x * x)/(0.1120-x) = 1.75 * 10^(-5)Ka=x⋅x0.1120−x=1.75⋅10−5
The value you'll get for xx is "0.0014"0.0014. Plug this into the pH equation and you'll get
pH_("initial") = -log(0.0014) = 2.85pHinitial=−log(0.0014)=2.85
Now you begin the titration by adding KOHKOH, which is a strong base. As a result, it will react with acetic acid, which is a weak acid, to form water and potassium acetate, a salt of acetic acid, according to this balanced equation
CH_3COOH_((aq)) + KOH_((aq)) -> CH_3COOK_((aq)) + H_2O_((l))CH3COOH(aq)+KOH(aq)→CH3COOK(aq)+H2O(l)
Notice that you have a "1:1"1:1 mole ratio between acetic acid and potassium hydroxide; this means that, at the equivalence point, the number of moles of potassium hydroxide used will be equal to the number of moles of acetic acid already in solution.
The number of moles of acetic acid you have in solution is
n_("acetic") = C * V = "0.1120M" * 47.32*10^(-3)"L" = "0.00530 moles"nacetic=C⋅V=0.1120M⋅47.32⋅10−3L=0.00530 moles
Automatically, both the acetic acid and the potassium hydroxide will be consumed in the reaction; the "1:1"1:1 mole ratio between either of these compounds and potassium acetate implies that you'll produce the same amount of CH_3COOKCH3COOK moles as you just consumed.
Now, you'll need to determine the total volume of the solution; for this you must determine how much KOHKOH was added
C = n/V => V_("KOH") = n/C = "0.00530 moles KOH"/"0.4750 M" = "11.16 mL"C=nV⇒VKOH=nC=0.00530 moles KOH0.4750 M=11.16 mL
The total volume of the solution will be
V_("total") = V_("acetic") + V_("KOH") = "47.25 mL" + "11.16 mL" = "58.41 mL"Vtotal=Vacetic+VKOH=47.25 mL+11.16 mL=58.41 mL
Use this volume to determine the concentration of the potassium acetate formed in solution
C = n/V_("total") = "0.00530 moles"/(58.41 * 10^(-3)"L") = "0.09074 M"C=nVtotal=0.00530 moles58.41⋅10−3L=0.09074 M
After the acetic acid and ptoassium hydroxide are consumed, CH_3COOKCH3COOK will react with water to form acetic acid and hydroxide anions; use the ICE chart again to solve for the concentration of hydroxide ions
.....CH_3COO_((aq))^(-) + H_2O_((l)) rightleftharpoons CH_3COOH_((aq)) + OH_((aq))^(-)CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)
I......0.09704..................................................0.......................0
C.....(-x)..........................................................(+x)...................(+x)
E...(0.09704 - x)............................................x.........................x
This time use the base dissociation constant, or K_bKb, which equals K_b = K_w/K_a = 10^(-14)/(1.75 * 10^(-5)) = 5.71 * 10^(-10)Kb=KwKa=10−141.75⋅10−5=5.71⋅10−10
Use this to determine xx
5.71 * 10^(-10) = x^(2)/(0.09704 - x) => x = 0.0000074445.71⋅10−10=x20.09704−x⇒x=0.000007444
Plug this baby into this equation to solve for the final pH
pH_("equiv") = 14 - pOH = 14 - log(0.000007444)pHequiv=14−pOH=14−log(0.000007444)
pH_("equiv") = 14 - 5.13 = 8.87pHequiv=14−5.13=8.87
