A solution is prepared by mixing 250 mL of 1.00 M CH3COOH with 500 mL of 1.00 M NaCH3COO. What is the pH of this solution? (Ka for CH3COOH = 1.8 × 10−5 )

A solution is prepared by mixing 250 mL of 1.00 M

CH3COOH with 500 mL of 1.00 M NaCH3COO.

What is the pH of this solution?

(Ka for CH3COOH = 1.8 × 10−5 )

Explanation:

This is a case of a neutralization reaction that takes place between acetic acid,     CH 3 COOH ,   a weak acid, and sodium hydroxide,   NaOH , a strong base.

The resulting solution pH, depends if the neutralization is complete or not.  If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.

It's important to note that at complete neutralization, the pH of the solution will not equal  7 . Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over  7 .

So, the balanced chemical equation for this reaction is the ionic equation:

CH 3 COOH (aq]  +  OH − (aq]  →  CH 3 COO − (aq]  +  H 2 O (l]

Notice that:  

1  mole of acetic acid will react with:  1  mole of sodium hydroxide, shown here as hydroxide anions,  OH − , to produce   1   mole of acetate anions:

CH 3 COO −

To determine how many moles of each you're adding , the molarities and volumes of the two solutions are used:

     c  =  n /  V    ⇒     n   =   c  ⋅  V

n  acetic   =   0.20 M   ⋅   25.00   ⋅   10  − 3 L   =   0.0050 moles CH3 COOH

and

n  hydroxide   =   0.10 M   ⋅   40.00   ⋅   10 − 3 L   =   0.0040 moles OH −

There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution is:

    n  acetic remaining   =   0.0050  −   0.0040   =    0.0010 moles

The reaction will also produce  0.0040   moles of acetate anions.

This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH :

pH  =  p K a  +  log  ( [ conjugate base ]  / [ weak acid ] )

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base .

V total  =  V acetic  +  V hydroxide

V total  =  25.00 mL  +  40.00 mL  =  65.00 mL

Thus the concentrations will be :

[ CH 3 COOH ]  =  0.0010 moles  / 65.00  ⋅  10 − 3 L  =  0.015385 M

and

[ CH 3 COO − ]  =  0.0040 moles  / 65  ⋅  10 − 3 L  =  0.061538 M

The    p K a     of acetic acid is equal to    4.75

Thus the pH of the solution will be:

pH   =   4.75  +  log ( 0.061538 M  /    0.015385 M )

pH   =   5.35

4.745

Explanation:

We are given from the question that 250 mL of 1.00 M CH3COOH and 500 mL of 1.00 M NaCH3COO makes a solution.

The equation of Reaction is given below;

CH3COOH + NaCH3COO > NaCH3COOH + CH3COONa.

==> Step 1 : Calculating the number of moles of each reactants and the find the limiting reagent.

One mole of CH3COOH reacts with one mole of NaCH3COO to give one NaCH3COOH and one mole CH3COONa.

Number of moles CH3COOH = Concentration of CH3COOH × volume of CH3COOH.

Number of moles CH3COOH= 1.00 M × 250 × 10^-3 litres.

Number of moles CH3COOH= 0.25 moles.

Number of moles of NaCH3COO = Concentration × volume.

Number of moles of NaCH3COO= 1.00 M × 500 × 10^-3 litres.

Number of moles of NaCH3COO= 0.5 moles.

Acetic acid is the limiting reagent. Therefore the numbers of moles of NaCH3COO that remains=( 0.5 - 0.25 )moles. = 0.25 moles.

=== > Step 2: use the formula in equation (1) below to calculate pH value.

First, the total volume of the solution =( 250 + 500)mL = 750 mL.

Thus, the Concentration of NaCH3COO= remaining moles/ total volume= 0.25 moles / 750 × 10^3 Litres.

Concentration of CH3COONa=0.25/ 0.75 = .33 M.

Concentration of CH3COOH = 0.25/ 0.75 = .33 M.

pH= pKa + log ([conjugate base]/ [weak acid] ) . (1).

pKa= - log ka.

pKa = - log 1.8 × 10^-5= 4.745.

Hence, pH= 4.745 + log ([.33)/ [.33]).

pH = 4.745 + 0.

pH = 4.745