Calculate the pH of a solution after 10.0 mL of 0.100 M NaOH is added tO₄0.0 mL of 0.250 M HBr.

Question

Quizplus · Accepted Answer

First we need to determine the moles of HBr before any reaction occurs.

moles HBr = (0.250 M) x (0.1000 L) = 0.0250 mol

Next, we need to determine the limiting reagent in the reaction.

NaOH + HBr → NaBr + H2O

The balanced equation shows a 1:1 mole ratio of NaOH to HBr. Therefore, the moles of NaOH added will be equal to the moles of HBr consumed.

moles NaOH = (0.100 M) x (0.0100 L) = 0.00100 mol

Since 0.00100 mol is less than 0.0250 mol, HBr is the limiting reagent.

The reaction between HBr and NaOH will result in the formation of NaBr and H2O. We need to determine how many moles of HBr are left over after the reaction.

moles HBr remaining = 0.0250 mol - 0.00100 mol = 0.0240 mol

Next, we can calculate the concentration of H3O+ in the solution using an ICE table.

HBr + OH- → H2O + Br-

I: 0.0240 mol/L     0.00100 mol/L        0 
C: -0.00100 mol/L  -0.00100 mol/L   +0.00100 mol/L 
E: 0.0230 mol/L    0.00000 mol/L       0.00100 mol/L

[H3O+] = 0.00100 mol/L

pH = -log[H3O+] = -log(0.00100) = 2.00

However, we need to take into account the fact that NaOH is a strong base, and will dissociate completely to form OH- when added to water.

NaOH → Na+ + OH-

Before any HBr is added, the solution contains only H2O and NaOH.

[OH-] = (0.100 M) x (0.0100 L) / (0.1000 L + 0.0100 L) = 0.00909 mol/L

pH = 14.00 - pOH = 14.00 - (-log[OH-]) = 14.00 - (-log 0.00909) = 0.74

Calculate the PH of a Solution After 10