Question #1bb81

ADDED INFO. Consider a weak acid, #HZ#, and its conjugate base, #Z^(-)#. The pH of a solution that is 1 M in #HZ# and 1M in #Z^(-)# is 5.0, so if I add .1M #NaOH# to that, pH would be?

The acid dissociation constant for #HZ# is #K_a = 1.0 * 10^(-5)#.

So, you're dealing with a buffer, which is a solution that contains a weak acid and its conjugate base (or a weak base and its conjugate acid).

The great thing about dealing with buffers is that you can use the Henderson-Hasselbalch equation to determine the pH of the solution.

#pH_"sol" = pK_a + log((["conjugate base"])/(["weak acid"]))#

The #pK_a# is actually the negative log of the acid dissociation constant, #K_a#

#pK_a = -log(K_a) = -log(1 * 10^(-5)) = 5#

When you have equal concentrations of weak acid and conjugate base, the pH of the solution will be equal to the #pK_a#

#pH_"sol" = pK_a + log(1) = pK_a = 5# #-># just like in your case.

Now you go ahead and add sodium hydroxide, a strong base. The sodium hydroxide will react with the weak acid

#HZ_((aq)) + NaOH_((aq)) -> NaZ_((aq)) + H_2O_((l))#

Since you didn't provide the volumes of the two solutions, I'll assume that you have 1 L of each.

The number of moles of weak acid would be

#C = n/V => n = C * V#

#n_(HZ) = "1 M" * "1 L" = "1 mole"#

For the sodium hydroxide, you have

#n_(NaOH) = "0.1 M" * "1 L" = "0.1 moles"#

The sodium hydroxide will react with the weak acid and get consumed. The number of moles of weak acid will decrease by how many moles of sodium hydroxide it reacted with, so

#n_("HZ left in sol") = n_(HZ) - n_(NaOH) = 1 - 0.1 = "0.9 moles"#

At the same time, you'll produce 0.1 moles of conjugate base, #Z^(-)#, so

#n_(Z^(-)"left in sol") = n_(Z^(-)) + 0.1 = 1 + 0.1 = "1.1 moles"#

The total volume of the solution will now be

#V_"total" = V_(HZ) + V_(NaOH) = 1 + 1 = "2 L"#

The concentrations of the weak acid and of its conjugate base will be

#[HZ] = "0.9 moles"/"2 L" = "0.45 M"#

#[Z^(-)] = "1.1 mole"/"2 L" = "0.55 M"#

Therefore, the pH of the solution will be

#pH_"sol" = pK_a + log(([Z^(-)])/([HZ]))#

#pH_"sol" = 5 + log(0.55/0.45) = color(green)(5.087)#

SIDE NOTE If you have other volumes for the solution, just use this answer as a model to guide you through the calculations.

I didn't take into account sig figs either.