Respuesta :

Both the acid and the base are known as strong substances which means they totally dissociate into ions in aqueous solution. Mixing the two a salt would be form which is NaNO3. The unreacted substances will be the one contributing to the pH of the solution. 

0.125M HNO3 x 0.04L = 0.005 mol HNO3
0.125M NaOH x 0.05L = 0.00625 mol NaOH

As we can see, all of the HNO3 would react with NaOH so the remaining OH from the unreacted NaOH would be responsible for the pH.

0.00625 - 0.005 / ( 0.0500L + 0.0400L ) = 0.0139 M 

pOH = -log( OH- ) 
pOH = 1.86 
pH = 14.00 - pOH 
pH = 12.14
okpalawalter8

The pH of a solution prepared is mathematically given as

pH = 12.14

What is the pH of a solution prepared?

Question Parameters:

The pH of a solution prepared by mixing 50.0 ml of 0.125 m NaOH and 40.0 ml of 0.125 m HNO3

0.125M HNO3 x 0.04L = 0.005 mol HNO3

0.125M NaOH x 0.05L = 0.00625 mol NaOH

Therefore, The Remainng OH for the reaction will have a molality of

0.00625 - 0.005 / ( 0.0500L + 0.0400L ) = 0.0139 M  

Generally, the equation for the pOH  is mathematically given as

pOH = -log( OH- )

pOH = 1.86

pH = 12.14

In conclusion,The ph of a solution prepared is

pH = 12.14

Read more about Chemical Reaction

https://brainly.com/question/11231920

Otras preguntas