A1.00 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. 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. 3.34 3.97 5.03 4.41 4.19

pH = 3.97

Explanation:

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

Note: We do need to calculate concentration since the volume is 1 L and do not affect the ratio   (A⁻) / (HA) and cancel out.

b.final

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error