pH question ...... //@.<) (1 Viewer)

[mecramarathon]

I can't seem to get the answer of pH 1.48

200mL of 0.1 mol/L NaOH is added to 100mL of 0.15 mol/L H*2SO*4. Assuming complete ionisation of the acid, the rseulting pH of thes olution is?



cheers to the person who can explain this with equations!

 

[annabackwards]

2NaOH + H2SO4 --> Na2SO4 + 2H2O

200mL of 0.1 mol/L NaOH is added to 100mL of 0.15 mol/L H*2SO*4.

Now n(NaOH) = cV = 200/1000 x 0.1 = 0.02 moles
n(H2SO4) = cV = 100/1000 x 0.15 = 0.015

Assuming complete ionisation of the acid, that means there are 0.015 x 2 = 0.03 moles of H+
Now there are 0.02 moles of OH- from the NaOH, so after neutralising there are 0.03 - 0.02 = 0.01 moles of extra H+

So the c[H+] = n/v = 0.01/(300/1000) = 0.03333333

Now pH = -log [H+] or pOH = -log [OH-]

Thus pH = -log [H+] = -log [0.03333333] = 1.477 = 1.48

 

[mecramarathon]

OMG thanks!!! :wave:

 

[annabackwards]

No worries ^^

 

[adomad]

don't we need to find the excess moles of H2SO4 first?