[biomed77]

hello everybody, i have problems with some calculations here and i was wondering if someone could help me out.  any help is much appreciated....

calculate the molarity of an aqueous solution of sodium hydroxide if 25cm^3 of it was neutralised by 40cm^3 of 0.05mol dm^-3 nitric acid. what i did is:  0.05*40/1000=0.002 and then 0.002/0.025=0.08M  is it right???

calculate the molarity of a hydrochloric acid solution if 20cm^3 of it was neutralised by 25cm^3 of 0.4mol dm^-3 aqueous sodium hydroxide.....0.4*0.025=0.01   0.01/0.020=0.5M  is it right?


and now am struggling to find the volume of 0.2mol dm^-3 hydrochloric acid that would be required to neutralise 20cm^3 of 0.3mol dm^-3 aqueous potassium hydroxide...HELP PLZ!!!

[Borek]

0.08M  is it right???

OK

0.5M  is it right?

OK

and now am struggling to find the volume of 0.2mol dm^-3 hydrochloric acid that would be required to neutralise 20cm^3 of 0.3mol dm^-3 aqueous potassium hydroxide...HELP PLZ!!!

Start with the balanced reaction equation.

[biomed77]

KOH+HCL= KCL+H20.....the equals sign should be an arrow and the 2 a small 2,sorry but dont know to use it very well. anyway what do i do next???/i did 0.3*20/1000= 0.006moles  and then isaid 0.2=0.006  because the ratio is 1:1  so 0.2/0.006= 33.3cm^3   but i am not sure.. thanks for responding anyway...

[Borek]

0.3*20/1000= 0.006moles

So far so good.

the ratio is 1:1

This one is OK too.

So you are left with the question "What volume of 0.2M HCl contains 0.006 moles of acid?".

And it is not 33.3 mL.

[biomed77]

20*0.3= 6  so 6/0.2= 30mL         is it right now?