A 400.0 ml sample of 0.10 mba(oh)2 is titrated with 0.10 mhbr. determine the ph of the solution before the addition of any hbr.

2 answers:

7 0

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

ikadub [295]4 years ago

4 0

The pH of the given 400 ml of 0.10 M ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ solution is $\boxed{{\text{13}}{\text{.3}}}$ .

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:

${\text{M}}=\frac{{{\text{n}}\left({{\text{mol}}}\right)}}{{{\text{V}}\left({\text{L}}\right)}}$ …… (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.

${\text{n}}\left({{\text{mol}}}\right)=\left({\text{M}}\right)\left({{\text{V}}\left({\text{L}}\right)}\right)$ …… (2)

The balanced dissociation reaction of barium hydroxide is,

${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}\to{\text{B}}{{\text{a}}^{2+}}+2{\text{O}}{{\text{H}}^-}$

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 ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ is 0.10 M and volume is 400.0 mL. Substitute these value in equation (2) to calculate the number of moles of ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ .

$\begin{aligned}{\text{n}}\left({{\text{mol}}}\right)&=\left({\text{M}}\right)\left({{\text{V}}\left({\text{L}}\right)}\right)\\&=\left({{\text{0}}{\text{.10}}\frac{{{\text{mol}}}}{{\text{L}}}}\right)\left({400.0{\text{ mL}}\times\frac{{1{\text{ L}}}}{{{{10}^3}{\text{ mL}}}}}\right)\\&=0.04{\text{ mol}}\\\end{aligned}$

Since 1 mole of ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ produces 2 moles of hydroxide ion, therefore, 0.04 moles of ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ will produce,

$\begin{aligned}{\text{Moles of O}}{{\text{H}}^-}&=2\left({{\text{Ba}}{{\left({{\text{OH}}}\right)}_{\text{2}}}}\right)\\&=2\times0.04\;{\text{mol}}\\&=0.08\;{\text{mol}}\\\end{aligned}$

Hence, concentration of ${\text{O}}{{\text{H}}^-}$ ions in the solution is calculated as follows:

$\begin{aligned}{\text{M}}&=\frac{{{\text{n}}\left({{\text{mol}}}\right)}}{{{\text{V}}\left({\text{L}}\right)}}\\&=\frac{{0.08{\text{ mol}}}}{{\left({400.0{\text{ mL}}\times\frac{{1{\text{ L}}}}{{{{10}^3}{\text{mL}}}}}\right)}}\\&=0.20\;{\text{M}}\\\end{aligned}$

The formula to calculate the pH of the solution is,

${\text{pH}}=14-{\text{pOH}}$

This expression can be elaborate as,

$\begin{aligned}{\text{pH}}&=14-{\text{pOH}}\hfill\\{\text{pH}}&=14-\left({-\log\left[{{\text{O}}{{\text{H}}^-}}\right]}\right)\hfill\\{\text{pH}}&=14+\log\left[{{\text{O}}{{\text{H}}^-}}\right]\hfill\\\end{aligned}$ …… (3)