Question

find the pH of a 40 mL buffer solution + 1.0 mL of 6.0 M HCl (aq)

I know this question is pretty simple, but do I need to know what the buffer is made of? Is it important? if not how would I set up the ICE table? thank you!

Answer 1

-Buffer is a mixture of weak acid and its conjugate base or weak base and its conjugate acid in aqueous solution.

-For a weak acid like acetic acid ,CH3COOH ,conjugate base is in the form of its salt like CH3COONa(sodium salt) and for a weak base like Ca(OH)2 ,ammonium hydroxide its conjugate acid would be in the form of its salt like CaCl2.

-Buffers shows very little change of pH (within the range of +/-1 on addition of a small amount of acid or base.

-Buffers are therefore used to maintain pH as a constant.

Let's see for a given buffer say CH3COOH+CH3COONa

Weak acid dissociates partly in aqueous solution,depending on its ka or acid dissociation constant

CH3COOH+H2O<---> CH3COO- +H3O+

ka=[ CH3COO-] [H3O+]/[CH3COOH]     [] represents concentration of the species within

salt dissociates completely, hence equilibrium exists as given in the above equation between acid and base

CH3COONa--------->CH3COO- +Na+

pH=pKa+log [base]/[acid]   given by such Henderson-hasselbach equation

or,pH=pKa+ log [CH3COO-]/[CH3COOH]

Now coming back to the problem "find the pH of a 40 mL buffer solution + 1.0 mL of 6.0 M HCl (aq)"

When strong acid like HCl is added it will neutralize its equal amount of conjugate base .

Amount of acid added =1.0 ml*6.0 mol/L=1/1000 L* 6.0 mol/L=6.0*10^-3 moles HCl added which neutralizes equal amount of base

remaining acid or base will be responsible for the pH next .

say 40 ml of buffer having 0.2M CH3COOH +0.2MCH3COONa

amount of base=0.2 mol/L *40 /1000 L=8.0*10^-3 moles

So after neutralization remaining base=8.0*10^-3 moles -6.0*10^-3 moles HCl=2.0 *10^-3 mol base (decreased) but acid amount increases by 2.0 *10^-3 mol.

HCl +CH3COO----------->CH3COOH+Cl- (acid is formed)

so pH=pka+ log (2.0 *10^-3 mol/8.0*10^-3 moles +2.0*10^-3 moles )

or,pH=pka -0.7

if ka is known pKa=-log ka

so pH can be calculated thus.