Chapter 14: Problem 58

Consider the titration of $80.0 \mathrm{mL}$ of $0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}$ by $0.400 M$ HCl. Calculate the $p H$ of the resulting solution after the following volumes of HCl have been added. a. $0.0 \mathrm{mL}$ b. $20.0 \mathrm{mL}$ c. $30.0 \mathrm{mL}$ d. $40.0 \mathrm{mL}$ e. $80.0 \mathrm{mL}$

Short Answer

Expert verified

a) $pH \approx 13.301$ b) $pH \approx 12.602$ Follow the same process for the remaining volumes to find the pH values for 30.0 mL, 40.0 mL, and 80.0 mL of HCl added.

Step by step solution

We must first find the initial amount of $\mathrm{Ba(OH)_2}$ in moles. To do this, multiply the initial volume by the molarity: $$moles = 80.0\ \mathrm{mL} \times 0.100\ \mathrm{M} = 0.00800\ \mathrm{mol}.$$ #Step 2: Determine moles and concentrations of reagent ions at each point of the titration# For each of the given HCl volumes, calculate the moles of HCl and find the prevailing concentrations of $\mathrm{OH^-}$ ions in the solution by accounting for the stoichiometry. If all of the $\mathrm{OH^-}$ ions are consumed, calculate the moles of $\mathrm{H_3O^+}$ ions generated. #Step 3: Calculate the pH at each stage of the titration# Use the concentrations of $\mathrm{OH^-}$ and $\mathrm{H_3O^+}$ ions to calculate the pH at every stage of the titration. Use the formula $pH = 14 - pOH$ if the $\mathrm{OH^-}$ ions haven't been consumed entirely and $pH = -\log_{10} [\mathrm{H_3O^+}]$ if you need to find the pH using the $\mathrm{H_3O^+}$ concentration. Now let's proceed to the specific calculation of the pH for each situation here: a) 0.0 $\mathrm{mL}$ HCl added #Step 1a: Determine the Initial Moles of $\mathrm{OH^-}$ ions#

Calculate the moles of $\mathrm{OH^-}$ ions from the initial moles of $\mathrm{Ba(OH)_2}$. As each mole of $\mathrm{Ba(OH)_2}$ dissociates into two moles of $\mathrm{OH^-}$ ions, we get: $$0.00800\ \mathrm{mol}\ \mathrm{Ba(OH)_2} \times 2\ \mathrm{mol}\ \mathrm{OH^-}/\mathrm{mol}\ \mathrm{Ba(OH)_2}=0.0160\ \mathrm{mol}\ \mathrm{OH^-}.$$ The concentration of $\mathrm{OH^-}$ ions is: $$\frac{0.0160\ \mathrm{mol}\ \mathrm{OH^-}}{80.0\ \mathrm{mL}} = 0.200\ \mathrm{M}.$$ #Step 2a: Calculate $pOH$ and $pH$#

Since the concentration of $\mathrm{OH^-}$ ions is known, we can find the $pOH$: $$pOH = -\log_{10} [\mathrm{OH^-}] = -\log_{10} (0.200) = 0.699.$$ Now, we can find the pH: $$pH = 14 - 0.699 = 13.301.$$ b) $20.0 \mathrm{mL}$ HCl added #Step 1b: Calculate the moles of $\mathrm{OH^-}$ ions and HCl#

Before adding the HCl, we know there are $0.0160\ \mathrm{mol}$ of $\mathrm{OH^-}$ ions. To find how many moles of HCl are added, multiply the volume by the concentration: $$20.0\ \mathrm{mL} \times 0.400\ \mathrm{M} = 0.00800\ \mathrm{mol}.$$ #Step 2b: Calculate the moles and concentrations of leftover ions#

In this case, two moles of HCl neutralize one mole of $\mathrm{Ba(OH)_2}$. After the reaction, there will be $0.00400\ \mathrm{mol}$ of $\mathrm{OH^-}$ ions left in the solution: $$0.0160\ \mathrm{mol} - 2 \times (0.00800\ \mathrm{mol}) = 0.00400\ \mathrm{mol}.$$ The $\mathrm{OH^-}$ concentration is: $$\frac{0.00400\ \mathrm{mol}}{80.0\ \mathrm{mL} + 20.0\ \mathrm{mL}} = 0.0400\ \mathrm{M}.$$ #Step 3b: Calculate $pOH$ and $pH$#

Since the concentration of $\mathrm{OH^-}$ ions is known, we can find the $pOH$: $$pOH = -\log_{10} [\mathrm{OH^-}] = -\log_{10} (0.0400) = 1.398.$$ Now, we can find the pH: $$pH = 14 - 1.398 = 12.602.$$ To find the pH in the other situations, follow the same process for 30.0 mL, 40.0 mL, and 80.0 mL of HCl added.