Need help, anyone good at chemistry perhaps?

[TechHelp]

Long shot but hey guys ive gotten pretty desparate with this question, if anyone can show me their working and the solution i'll be forever appreciative

"Calculate the final pH of the resultant solution if: 5.0g of sodium hydroxide were completely dissolved in 120mL of distilled water and this solution was added to 200mL of 0.2mol.L^-1 nitric acid."

 

thank you

[TechHelp]

Just now, James Evens said:

What did you already come up with?

Easily got the equation, got moles of everything and concentration but it's just the pH part I'm struggling with im pretty sure I can find the pH of the individual components but no clue for the resultant solution

 

[TechHelp]

Just now, James Evens said:

strong or weak acid/base?

what is the definition of ph?

Sodium hydroxide strong base and nitric acid strong acid

And pH is a measure of acidity??

 

[dgsddfgdfhgs]

HNO3 + NaOH = NaNO3 + H2O

you need to find out which one is in excess after reaction, h / oh

[TechHelp]

Just now, dgsddfgdfhgs said:

HNO3 + NaOH = NaNO3 + H2O

you need to find out which one is in excess after reaction, h / oh

How does that help when finding final pH though??

 

[dgsddfgdfhgs]

6 minutes ago, TechHelp said:

How does that help when finding final pH though??

this part wasnt in my syllabus, but from web, theres some formula of doing so with log fn

 

1st random link:

https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm

[davrosG5]

1 hour ago, TechHelp said:

Sodium hydroxide strong base and nitric acid strong acid

And pH is a measure of acidity??

 

Both components are 'strong' so you don't have to worry about the ionisation potential of either - it would be different if you had something like acetic acid or ammonia for example.

 

pH is the -log (to the base 10) of the Hydrogen ion concentration

It's counterpart, pOH is the -log (to the base 10) of the Hydroxide ion concentration.

The sum of pH and pOH is 14 so once you know the concentration of one of the ions you can work out the rest.

 

You said you had worked out the concentrations - the acid will have been completely neutralised as the NaOH was in excess. Whatever your final NaOH concentration is also the OH concentration so you can calculate the pOH value then subtract it from 14 to get the pH. I got a pH of ~13.4.

[davrosG5]

Okay:

5g NaOH = 5/40 moles = 0.125 moles NaOH (in 120mL = 1.042 moles/L)

200 mL of 0.2M HNO3 = 0.2moles/L x 0.2L = 0.04 moles HNO3

You can ignore the Na+ and NO3- counter ions.

H+ + OH- -> H2O

The OH- is in excess so effectively 'all' of the H+ is neutralised so your final OH- concentration is 0.125 - 0.04 moles = 0.085 moles

You have 0.085 moles of OH- in 320mL of water so you [OH-] = 0.085 moles/0.32L = 0.265625 moles/L

 

pOH = -log10 (0.265625) = 0.5757

 

14 = pH + pOH

pH = 14 - pOH = 14 - 0.5757 = 13.4243 or about 13.4