posted by auro .
determine the quantity of NaOH whose concentration is 15% w/v to neutralize 176g H2SO4 , this is the exact problem and no its not an experiment. sorry to bother someone of you
Thanks. I can help with this. And the experiment part of my question meant that if an experiment I would need to know what you were doing.
As you have written it the problem now makes sense.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
First, how many moles NaOH do you have in the solution. That is 15% w/v which means 15 g NaOH/100 mL solution. moles = grams/molar mass = 15/40 = about 0.38 moles NaOH. You can redo this and be more exact. So the 15% w/v NaOH is about 0.38 moles/100 mL.
176 g H2SO4 is how many moles.
176/98 = about 1.8 moles. Convert moles H2SO4 to moles NaOH using the coefficients in the balanced equation. That will be 1.8 moles H2SO4 x (2 moles NaOH/1 mole H2SO4) = 1.8 x (2/1) = about 3.6 moles NaOH. Our NaOH has about 0.38 moles/100 mL, so how many mL will it take to obtain 3.6 moles. That will be 100 mL x (3.6 moles/0.38 moles) = about 947 mL of the 15% NaOH solution will be required. .
thanks a lot