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What would be the final ph if 0.0100 moles of solid naoh were added to 100ml of a buffer solution containing 0.600 molar formic acid (ionization constant = 1.8x10-4) and 0.300 m sodium formate?

Respuesta :

superman1987
moles of formic acid = 0.6 M * 0.1L = 0.06 m
moles of sodium formate = 0.3 M * 0.1L = 0.03 m
According to the reaction equation:
 
                     HCOOH(aq) + NaOH(s) ↔ HCOO- (aq) + H2O(l)
intial             0.06                  0.01               0.03
change       - 0.01                 -0.01               +0.01
final             0.05                    0                   0.04

So when :
PH =PKa + ㏒(strong base/weak acid)
when we have Ka =1.8x10^-4 and sodium formate is the strong base and formic acid is the weak acid.
PH = -㏒ Ka + ㏒(0.04/0.05)
      = -㏒1.8x10^-4 + ㏒ (0.04/0.05)
 ∴PH  =3.65

BladeRunner212

3.65

Further explanation

Given:

  • 100 ml of a buffer solution containing 0.600 molar formic acid and 0.300 m sodium formate.
  • The Ka for formic acid is 1.8 x 10⁻⁴.

Question:

What would be the final ph if 0.0100 moles of solid NaOH were added to this buffer?

The Process:

Step-1

Let us prepare all the moles of substances.

[tex]\boxed{ \ n = MV \ }[/tex]  

Moles of HCOOH =  

[tex]\boxed{ \ 0.600 \ \frac{mol}{L} \times 100 \ ml = 60 \ mmol \ }[/tex]

Moles of HCOONa =  

[tex]\boxed{ \ 0.300 \ \frac{mol}{L} \times 100 \ ml = 30 \ mmol \ }[/tex]

Moles of NaOH =

[tex]\boxed{ \ 0.0100 \ moles = 10 \ mmol \ }[/tex]

Step-2

Let use the ICE table (in mmol).

                  [tex]\boxed{ \ HCOOH_{(aq)} + NaOH_{(s)} \rightarrow HCOONa_{(aq)} + H_2O_{(l)} \ }[/tex]

Initial:                  60                  10                   30                   -

Change:             -10                 -10                 +10                +10

Equlibrium:       50                   -                    40                  10

NaOH as a strong base acts as a limiting reagent.

The remaining HCOOH as a weak acid and HCOONa salt forms an acidic buffer system.

The HCOONa salt has valence = 1 according to the number of HCOO⁻ ions as a weak part, i.e.,  [tex]\boxed{ \ HCOONa \rightleftharpoons HCOO^- + Na^+ \ }[/tex]

HCOOH and HCOO⁻ are conjugate acid-base pairs.

Step-3

To calculate the specific pH of a given buffer, we need using The Henderson-Hasselbalch equation for acidic buffers:

[tex]\boxed{ \ pH = pK_a + log\frac{[A^-]}{[HA]} \ }[/tex]

where,  

  • Ka represents the dissociation constant for the weak acid;
  • [A-] represent the concentration of the conjugate base (i.e. salt);  
  • [HA] is the concentration of the weak acid.

[tex]\boxed{ \ pH = pK_a + log\frac{[HCOO^-]}{[HCOOH]} \ }[/tex]

[tex]\boxed{ \ pH = -log(1.8 \times 10^{-4}) + log \Big(\frac{40}{50}\Big) \ }[/tex]

[tex]\boxed{ \ pH = 4 - log \ 1.8 - 0.0969 \ }[/tex]

[tex]\boxed{ \ pH = 4 - 0.2553 - 0.0969 \ }[/tex]

[tex]\boxed{ \ pH = 3.65 \ }[/tex]

Thus, the pH of this buffer equal to 3.65.

_ _ _ _ _ _ _ _ _ _

What if we calculate the buffer pH value before the addition of NaOH?

Moles of HCOOH =  60 mmol

Moles of HCOONa =  30 mmol

[tex]\boxed{ \ pH = pK_a + log\frac{[HCOO^-]}{[HCOOH]} \ }[/tex]

[tex]\boxed{ \ pH = -log(1.8 \times 10^{-4}) + log \Big(\frac{30}{60}\Big) \ }[/tex]

[tex]\boxed{ \ pH = 4 - log \ 1.8 - 0.301 \ }[/tex]

[tex]\boxed{ \ pH = 4 - 0.2553 - 0.301 \ }[/tex]

[tex]\boxed{ \ pH = 3.44 \ }[/tex]

Thus, the initial pH of this buffer is 3.44. This proves the nature of the buffer that keeps the pH value relatively unchanged with the addition of a strong electrolyte.

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