A 1.00 L buffer solution is 0.150 M in HC7H5O2 and 0.250 M in LiC7H5O2. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HC7H5O2 is 6.5 × 10-5.

pH = 3.97

Explanation:

It looks like HC₇H₅O₂ is benzoic acid   C ₆ H ₅ C O O H

For short - hand I'll call this  H A

Here we have to determine the pH of a buffer solution. The Henderson-Hasselbalch give us the pH through the equation:

pH = pKa + log ((A⁻) / (HA))

where pKa  is the -log Ka,

         (A⁻)   is the concentration of the conjugate base of the weak acid

         (HA)  is the concentration of the weak acid

By adding the strong acid HCl  to the buffer, some of the conjugate base of the weak acid will be consumed and some of the weak acid will produced through the chemical equation:

HCl +  LiC₇H₅O₂   ⇒   HC₇H₅O₂  + LiCl

Therefore we need to account for this reaction to answer the question:

Vol HCl = 100 mL = 0.100 L

mol HCl reacted = 0.100 L x 1.00 mol/L = 0.100 mol

mol  LiC₇H₅O₂ originally present = 0.250 mol/L x 1.00  L = 0.250 mol

mol LiC₇H₅O₂ after reaction with HCl = 0.250 -0.100 M = 0.250 mol

mol HC₇H₅O₂   originally present = 0.150 mol/L x 1.00 L  = 0.150 mol

mol HC₇H₅O₂  produced in reaction = 0.100 mol

mol  HC₇H₅O₂   present after reaction = 0.150 mol + 0.100 mol = 0.250 mol

Now we are in position to calculate the pH by plugging our values into the equation:

pH = pKa + log ((A⁻) / (HA)) = 4.20 + log (0.15/0.25) = 4.20 + log 0.60

= 3.97

pH = 3.97

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.