Respuesta :
Ba(OH)2 is an basic solution. It has more OH- ions than H+ ions. pOH should be calculated to find out its pH
The reaction is
Ba(OH)2 ⇒ Ba2+ (aq) + 2 OH-(aq)
One mole barium hydroxide releases 2 moles hydroxide ions.
Use that ratio to calculate molarity (M) of OH- ions [OH-]. The ratio is 1:2.
0.10 M Ba(OH)2 release 2*0.10 M= 0.02 M OH- ions
[OH-]= 0.02
pOH= - log [OH-] = - log 0.02 = 1.7
Thats not the answer! We found pOH of the solution before titration.
pH and pOH relationship is shown by formula of pH+pOH= 14
pH= 14-pOH
pH= 14-1.7= 12.3
The reaction is
Ba(OH)2 ⇒ Ba2+ (aq) + 2 OH-(aq)
One mole barium hydroxide releases 2 moles hydroxide ions.
Use that ratio to calculate molarity (M) of OH- ions [OH-]. The ratio is 1:2.
0.10 M Ba(OH)2 release 2*0.10 M= 0.02 M OH- ions
[OH-]= 0.02
pOH= - log [OH-] = - log 0.02 = 1.7
Thats not the answer! We found pOH of the solution before titration.
pH and pOH relationship is shown by formula of pH+pOH= 14
pH= 14-pOH
pH= 14-1.7= 12.3
The pH of the given 400 ml of 0.10 M [tex]{\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}[/tex]solution is[tex]\boxed{{\text{13}}{\text{.3}}}[/tex].
Further explanation:
Molarity:
The molarity of the solution can be defined as the concentration of the solution and is equal to the number of moles of the solute dissolved in 1 liter of the solution.
The expression of molarity (M), volume (V), and a number of moles (n) is as follows:
[tex]{\text{M}}=\frac{{{\text{n}}\left({{\text{mol}}}\right)}}{{{\text{V}}\left({\text{L}}\right)}}[/tex] …… (1)
Here, V is a volume of solution in liters and n is a number of moles of solute.
Rearrange the above equation to calculate the number of moles from molarity.
[tex]{\text{n}}\left({{\text{mol}}}\right)=\left({\text{M}}\right)\left({{\text{V}}\left({\text{L}}\right)}\right)[/tex] …… (2)
The balanced dissociation reaction of barium hydroxide is,
[tex]{\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}\to{\text{B}}{{\text{a}}^{2+}}+2{\text{O}}{{\text{H}}^-}[/tex]
Since barium hydroxide is a strong electrolyte. Therefore, according to the balanced equation,1 mole of barium hydroxide completely dissociates into 1 mole of barium ion and 2 moles of hydroxide ion in aqueous solution.
Given molarity of [tex]{\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}[/tex] is 0.10 M and volume is 400.0 mL. Substitute these value in equation (2) to calculate the number of moles of [tex]{\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}[/tex].
[tex]\begin{aligned}{\text{n}}\left({{\text{mol}}}\right)&=\left({\text{M}}\right)\left({{\text{V}}\left({\text{L}}\right)}\right)\\&=\left({{\text{0}}{\text{.10}}\frac{{{\text{mol}}}}{{\text{L}}}}\right)\left({400.0{\text{ mL}}\times\frac{{1{\text{ L}}}}{{{{10}^3}{\text{ mL}}}}}\right)\\&=0.04{\text{ mol}}\\\end{aligned}[/tex]
Since 1 mole of [tex]{\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}[/tex] produces 2 moles of hydroxide ion, therefore, 0.04 moles of [tex]{\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}[/tex] will produce,
[tex]\begin{aligned}{\text{Moles of O}}{{\text{H}}^-}&=2\left({{\text{Ba}}{{\left({{\text{OH}}}\right)}_{\text{2}}}}\right)\\&=2\times0.04\;{\text{mol}}\\&=0.08\;{\text{mol}}\\\end{aligned}[/tex]
Hence, concentration of [tex]{\text{O}}{{\text{H}}^-}[/tex]ions in the solution is calculated as follows:
[tex]\begin{aligned}{\text{M}}&=\frac{{{\text{n}}\left({{\text{mol}}}\right)}}{{{\text{V}}\left({\text{L}}\right)}}\\&=\frac{{0.08{\text{ mol}}}}{{\left({400.0{\text{ mL}}\times\frac{{1{\text{ L}}}}{{{{10}^3}{\text{mL}}}}}\right)}}\\&=0.20\;{\text{M}}\\\end{aligned}[/tex]
The formula to calculate the pH of the solution is,
[tex]{\text{pH}}=14-{\text{pOH}}[/tex]
This expression can be elaborate as,
[tex]\begin{aligned}{\text{pH}}&=14-{\text{pOH}}\hfill\\{\text{pH}}&=14-\left({-\log\left[{{\text{O}}{{\text{H}}^-}}\right]}\right)\hfill\\{\text{pH}}&=14+\log\left[{{\text{O}}{{\text{H}}^-}}\right]\hfill\\\end{aligned}[/tex] …… (3)
Substitute [tex]0.20{\text{ M}}[/tex] for [tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex]in equation (3)
[tex]\begin{aligned}{\text{pH}}&=14+\log\left[{{\text{O}}{{\text{H}}^-}}\right]\\&=14+\log\left({0.20}\right)\\&=13.3\\\end{aligned}[/tex]
Therefore, the pH of the given solution is 13.3.
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Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Acid, bases and salts
Keywords: pH, 0.10 M baoh2, solution of barium hydroxide, pH 13.3, 2 mole oh- ions.
