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A solution of naoh(aq) contains 6.6 g of naoh(s) per 100.0 ml of solution. calculate the ph and the poh of the solution at 25 °c.

Respuesta :

kenmyna
The  POh  and  Ph  of  the  solution  at  25 degrees is  calculated  as  follows

The  moles   of  the   solution =  6.6g/40(molar mass  of  NaOH)= 0.165 moles
concentration=  ( 0.165/100)  x1000= 1.65 M
NaOH in  solution  dissociated   as  follows
NaOH  --->  Na+    +  OH-
  POh= -  log  (OH-)

POh=  -log(1.65) =  -0.22

Ph=14-POh
that is  14 --0.22=14.22

The pH and pOH of 6.6g of NaOH in 100mL solution are -0.217 and 14.217 respectively

Data;

  • Volume  = 100.0mL
  • mass of NaOH = 6.6g
  • molarity of NaOH = ?

Molarity of the Solution

To find the pOH and pH of this solution, we have to know the molarity of this solution.

Molarity = number of moles of solute / volume of the solution

number of moles of the solute = mass / molar mass

molar mass of NaOH = 40g/mol

number of moles = 6.6/40 = 0.165moles

Molarity of this solution is

[tex]M = \frac{number of moles }{volume of solution}\\M = \frac{0.165}{0.1}\\ Molarity = 1.65M[/tex]

pOH of the Solution

[tex]pOH= -log[OH^-]\\pOH = -log[1.65]\\pOH = -0.217[/tex]

At 25°C, the pOH of NaOH is -0.217, let's calculate the pH

[tex]pOH+pH=14\\-0.217+pH=14\\pH=14-(-0.217)\\pH=14.217[/tex]

From the calculations above, the pH and pOH of 6.6g of NaOH in 100mL solution are -0.217 and 14.217 respectively.

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