Consider the following balanced reaction. HCl (aq) + KOH (aq) → H2O (l) + KCl (aq)

Question

In a beaker, 250.0 mL of 0.25 M HCl and 200.0 mL of 0.40 M KOH are mixed.

1. After reaction, is the solution acidic, neutral, or basic?
2. What is the pH of the solution at the end of the reaction? Report answer to 4 significant figures.

Answer 1

HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)

Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:

moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl

moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH

1). You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC

2). To find the final pH, we will want to find how much excess OH- there is from the KOH. Once we know that, we can find the pOH and then then pH. There are other ways to do this, but this seems the easiest.

-- The excess KOH will be the 0.0800 moles KOH minus the 0.06250 moles KOH used to neutralize the HCl, thus this = 0.0800 - 0.06250 = 0.01750 moles KOH

-- Concentration of OH- will be 0.01750 moles KOH/0.450 L because the volume is now 250 ml + 200 ml = 450 ml = 0.450 L. So, final [OH-] = 0.01750/0.450 = 0.03889 M

-- pOH = -log [OH-] = -log 0.03889 = 1.410

pH = 14 - pOH

pH = 14 - 1.410

pH = 12.5898 (to 4 sig. figs.) (note: when taking log of a number, sig figs are in the mantissa, the # to the right of the decimal point)