# Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH solution...

## Question:

Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH solution after the addition of 54.00 mL of the NaOH solution.

## Strong-Strong Acid-Base Titration:

An acid-base titation reacts an acidic/basic analyte reactant with a basic/acidic titrant reactant to create two products. If both reactants are relatively strong, then the reaction solution pH is simple to predict, since neither of the products will shift the solution pH from neutrality. Therefore, the equivalence point of a strong-strong titration will always have a neutral pH. If you are above or below the equivalence point, then the pH will quickly deviate into a strongly acidic or basic region.

HCl is a strong monoprotic acid analyte reactant. NaOH is a strong monoprotic base titrant. The balanced reaction equation is:

{eq}\rm NaOH + HCl \rightarrow H_2O + NaCl {/eq}

First we find the starting molar quantities from the given volume and molarity data:

Before reaction:

• Moles of NaOH = {eq}\rm V \times C = 0.05400 \ L \times 0.1000 \ M = 0.005400 \ mol {/eq}
• Moles of HCl = {eq}\rm V \times C = 0.04000 \ L \times 0.1000 \ M = 0.004000 \ mol {/eq}

NaOH is in excess. Its remaining molarity will completely dissociate to provide the same molarity of hydroxide ion. This will create a basic pH in the final total volume present after reaction :

After reaction:

• Moles of NaOH remaining = {eq}\rm 0.005400 - 0.004000 = 0.001400 \ mol {/eq}
• Final volume = {eq}\rm 0.05400 + 0.04000 = 0.09400 \ L {/eq}
• {eq}\rm [OH^-] = \dfrac{0.001400 \ mol}{0.09400 \ L} = 0.01489 \ M {/eq}
• {eq}\rm pOH = -log[OH^-] = -log(0.01489 \ M) = 1.827 {/eq}
• {eq}\rm pH = 14.000 - 1.827 = 12.173 {/eq}