In: Chemistry
Part B
What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.
Express the pH numerically to three decimal places.
Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.507 mol of NaA in 2.00 L of solution? The dissociation constant K a of HA is 5.66× 10 −7 . Express the pH numerically to three decimal places
PART A
First we need to calculate molar concentrations of weak acid & its salt.
We have, [ HA ] = No. of moles of HA / volume of solution in L
[ HA ] = 0.809 mol / 2.00 L = 0.4045 M
[ NaA ] = 0.507 mol / 2.00 L = 0.2535 M
pH of buffer solution is calculated by using Henderson's equation.
pH = pKa + log [ NaA ] / [ HA ]
pH = - log Ka + log [ NaA ] / [ HA ]
pH = - log ( 5.66 10 -07 ) + log 0.2535 / 0.4045
pH = 6.247 + log 0.2535 / 0.4045
pH = 6.247 - 0.2029
pH = 6.044
PART B
Consider reaction of HCl with buffer solution.
When HCl is added to buffer solution. It reacts with basic component of buffer solution . In this case, HCl reacts with NaA & produces HA. Therefore, concentration of HA increases & that of NaA decreases.
NaA + HCl HA + NaCl
Let's use ICE table.
Concentration ( moles ) | NaA | HCl | HA |
I | 0.507 | 0.150 | 0.809 |
C | - 0.150 | -0.150 | +0.150 |
E | 0.357 | 0.000 | 0.959 |
pH of buffer solution after addition of HCl is calculated by using equation pH = pKa + log [ NaA ] / [ HA ]
pH = 6.247 + log ( 0.357 mol / 2.00 L ) / ( 0.959 mol / 2.00 L )
pH = 6.247 - 0.4292
pH = 5.818
ANSWER : pH buffer solution after addition of HCl is 5.818