Question

Calculate the pH value obtained when 1.0 mL of 0.1 M HCl is added to 99.0

mL of phosphate buffer containing 0.063 M HPO4

2- and 0.10 M H2PO4-

(Ratio 0.63:1).​

Answer 1

Explanation:

pH of buffer after addition of 1 mL of 0,1 MHCI = 7,0

Explanation:

It is possible to use Henderson-Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + logo

Where A is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H2PO4 2 HPO42- + H+ Kaz = 6,20x10-8; pka=7,2

Thus, Henderson-Hasselbalch equation for 7,00

phosphate buffer is:

HP04 7.0 = 7,2 + log10 H2POA|

Ratio obtained is:

0,63 = HPO4 [H2PO4 ]

As the problem said you can assume (H2PO4) = 0,1 M and [HPO42-1 = 0,063M

As the amount added of HCl is 0,001 M the concentrations in equilibrium are:

H2PO4 HPO42- + H*

0,1 M +x 0,063M -x 0,001M -x -because the

addition of H* displaces the equilibrium to the

Knowing the equation of equilibrium is:

Ka = HPo ||H*] (H,PO,

Replacing:

6,20x10-8 = 10,063-z|(0,001-] 10,1+|

You will obtain:

x -0,064 x + 6,29938x10-5 = 0

Thus:

x = 0,063 - No physical sense

x = 0,00099990

Thus, [H*] in equilibrium is:

0,001 M - 0,00099990 = 1x10-7

Thus, pH of buffer after addition of 1 mL of 0,1 M HCI

-log1o [1x10-) = 7,0

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!

I hope it helps!