Question

Assuming complete dissociation, what is the ph of a 4.02 mg/l ba(oh)2 solution?

Answer 1

Answer: The pH of the solution is 9.68

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Mass of solute (barium hydroxide) = 4.02 mg = 0.00402 g   (Conversion factor:  1 g = 1000 mg)

Molar mass of barium hydroxide = 171.34 g/mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{0.00402g}{171.34g/mol\times 1L}\\\\\text{Molarity of solution}=2.4\times 10^{-5}M$

To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

On complete dissociation, 1 mole of barium hydroxide produces 2 moles of hydroxide ions

We are given:

$[OH^-]=4.8\times 10^{-5}M$

Putting values in above equation, we get:

$pOH=-\log(4.8\times 10^{-5})\\\\pOH=4.32$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4.32=9.68$

Hence, the pH of the solution is 9.68

Answer 2

Ba(OH)2   dissociate   according  to   the  following  equation
Ba(OH)2---->  Ba 2+  +  2OH-
Moles  of  Ba(OH)2  =  4.02 /171(molar  mass  of  Ba(OH)2 )=  0.0235   moles
by  use  of  mole  ratio  of  Ba(OH)2  to  OH-   which  is 1:2  the  moles  of  OH =  0.0235  x  2=0.047  moles  in  one  liter   =  0.047/1000=  4.7  x10^-5
POh=-  log  (OH-)
that  is   -log (4.7  x10^-5)  =  4.34
Ph=  14  - POH
PH  is  therefore =  14  - 4.34 = 9.66