21.00 mL of 0.200 M HBr is titrated with 0.351 M NaOH.

What is the pH after the addition of 6.56 mL of the base ?

HBr + NaOH ==> H2O + NaBr

moles HBr initially = M x L = ??
moles NaOH added = M x L = ??
Whis # mols is larger; that will be the one in excess and (H^+) or (OH^-) (depending upon the one in excess) will equal to the concn of the reagent in excess.

i still cant make sense of this

To determine the pH after the addition of 6.56 mL of the base, we need to calculate the resulting concentration of HBr and NaOH, and then use this information to calculate the concentration of H+ ions in the solution.

Step 1: Calculate the moles of HBr initially present.
To do this, we use the formula:

moles of solute = concentration x volume
moles of HBr = 0.200 M x 21.00 mL
moles of HBr = 0.0042 mol

Step 2: Calculate the moles of NaOH added.
Again, we use the formula:

moles of solute = concentration x volume
moles of NaOH = 0.351 M x 6.56 mL
moles of NaOH = 0.00230 mol

Step 3: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of HBr and NaOH. The reactant that produces fewer moles is the limiting reactant. In this case, NaOH is the limiting reactant because it produces fewer moles (0.00230 mol) compared to the moles of HBr (0.0042 mol).

Step 4: Calculate the excess moles of HBr.
To find the excess moles of HBr, subtract the moles of NaOH from the initial moles of HBr.

excess moles of HBr = initial moles of HBr - moles of NaOH
excess moles of HBr = 0.0042 mol - 0.00230 mol
excess moles of HBr = 0.0019 mol

Step 5: Calculate the new concentration of HBr.
To find the new concentration of HBr, divide the excess moles by the new volume of the solution.

new concentration of HBr = excess moles of HBr / new volume of solution
new volume of solution = initial volume of HBr + volume of NaOH
new volume of solution = 21.00 mL + 6.56 mL
new volume of solution = 27.56 mL

new concentration of HBr = 0.0019 mol / 27.56 mL
new concentration of HBr = 0.0689 M

Step 6: Calculate the concentration of OH- ions.
Since NaOH completely dissociates in water, the concentration of OH- ions is equal to the concentration of NaOH.

concentration of OH- ions = concentration of NaOH
concentration of OH- ions = 0.351 M

Step 7: Calculate the concentration of H+ ions.
To calculate the concentration of H+ ions, we use the equation for neutralization:

HBr + NaOH -> NaBr + H2O

Since the reaction is 1:1, the change in concentration of H+ ions is equal to the concentration of OH- ions.

concentration of H+ ions = concentration of OH- ions
concentration of H+ ions = 0.351 M

Step 8: Calculate the pH.
To find the pH, we use the equation:

pH = -log10 [H+]
pH = -log10 (0.351)

Using a calculator, we find that pH = 0.455