A 25.0 mL sample of a 0.100 M solution of acetic acid is titrated with a 0.125 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. The Ka for acetic acid is 1.76×10–5.

Ater adding 10.0 mL the pH = 4.74

After adding 20.0 mL the pH = 8.75

After adding 30.0 mL the pH = 12.36

Explanation:

Step 1: Data given

Volume of a 0.100 M acetic acid solution = 25.0 mL

Molarity of NaOH solution = 0.125 M

A) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH

Calculate moles

Equation:   NaOH    +   HAc    -->   NaAc + H20

Moles HAc =  0.025 L*  0.100 mol /L = 0.00250 moles

Moles NaOH = 0.01 L * 0.125mol/L = 0.00125 moles

Initial moles

NaOH = 0.00125 moles

HAc = 0.00250 moles

NaAC = 0 moles

Moles at the equilibrium

NaOH = 0.00125 - 0.00125 = 0 moles

HAc = 0.00250 - 0.00125 = 0.00125 moles

NaAC = 0.00125 moles

Calculate molarity

[HAc] = [Ac-] =  0.00125 moles / 0.035 L = 0.0357M

Concentration at equilibrium

HAc   <--> Ac-     +  H+         Ka = 1.76 * 10^-5

[HAc] = 0.0357 - x M

[Ac-] = 0.0357 + X M

[H+] = XM

Ka=1.76*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]

Ka=1.76*10^-5 = [0.00357][x]/[0.0357] usually good approx.

Ka = 1.76* 10-5 = x = [H+]

pH = -log [H+] = -log [1.76*10^-5] = 4.74

B) 25.0mL HAc and 20.0mL NaOH

Calculate moles

Equation:   NaOH    +   HAc    -->   NaAc + H20

Moles HAc =  0.025 L*  0.100 mol /L = 0.00250 moles

Moles NaOH = 0.02 L * 0.125mol/L = 0.00250 moles

Initial moles

NaOH = 0.00250 moles

HAc = 0.00250 moles

NaAC = 0 moles

Moles at the equilibrium

NaOH = 0.00250 - 0.00250 = 0 moles

HAc = 0.00250 - 0.00250 = 0 moles

NaAC = 0.00250 moles

Calculate molarity

[NaAC] =  0.00250 moles / 0.045 L = 0.0556M

Concentration at equilibrium

Ac- + H20   <--> HAc     +  OH-        Ka = 1.76 * 10^-5

[Ac] = 0.0556 - x M

[OH-] = XM

[HAc] = XM

**Kw = 1*10^-14 at STP I'll assume this for this problem

**Kw = Ka*Kb ⇒ Kb = Kw/Ka

**Kb = (1*10^-14)/(1.76*10^-5)= 5.68*10^-10  

Kb = ([HAc][OH-])/[Ac-] =  ([x][x])/[0.0556]

Kb =  x² /0.0556

5.68*10^-10 = x² /0.0556

so x =5.62*10^-6

pOH = -log [OH-] = 5.250

pH = 14 - pOH

pH 14 - 5.25 = 8.75

C)25.0mL HAc and 30.0mL NaOH

NaOH    +   HAc    -->   NaAc + H20

Initial numbers of moles

Moles NaOH = 0.125M * 0.03 L = 0.00375 moles

Moles HAc =  0.00250 moles

moles NaAC = 0 moles

Moles at the equilibrium

Moles NaOH = 0.00375 moles - 0.00250 = 0.00125 moles

Moles HAc =  0.00250 moles - 0.00250 = 0 moles

moles NaAC = 0.00250 moles

After the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.

Calculate NaOH molarity

(0.00125moles/0.055 L= 0.0227M NaOH

Strong Bases dissociate 100% so [OH-] = 0.0227M

pOH= -log[0.0227] = 1.644

pH = 14-pOH = 12.36

answer: option (4) is the correct answer.

explanation:

areas which are at low altitudes and near the equator tend to receive direct sunlight from the sun. hence, heat energy or radiations from the sun make these areas warmer and they have warmest water.

whereas areas which are at high altitudes and away from the equator do no receive direct sunlight. hence, no direct contact from the sun makes these regions colder and thus, they have colder water.

thus, we can conclude that the statement areas at low latitudes near the equator have the warmest water because they receive the most direct sunlight best explains the relationship between an area's geography and the temperature of its surface water.

the answers would be

the first one and the second one