Chemistry help?
What volume of 0.5 M nitric acid is required to neutralize 25 ml of a 0.2 M sodium hydroxide solution?
a. write the balanced equation.
b.since the concentration of the base is given, determine the moles of the base.
c. from the coefficients of the balanced equation, simplify for the mole(s) of acid required to neutralize the base.
d. solve for the volume of the acid.
I don't understand this...if you could help me out, or explain, or preferably both:) I would be forever grateful...thank you!
Answer:
a.) HNO3 + NaOH = NaNO3 + H2O
b.) Moles base = 25 x 0.2 /1000 = 0.005 moles
0.2M means 0.2 moles sodium hydroxide (NaOH) per liter(1000ml) so 25 ml contains 0.005 moles
c.) 1 mole acid is required for 1 mole of base
d.) Volume of acid = 25 x0.2 / 0.5 = 10 ml
The acid is 2.5 x stronger than the base so only need 10ml.
Vol acid x molarity acid = vol base x molarity base
i.e 10 x .5 = 25 x 0.2
This question is about an acid-base titration:
An acid aka proton donor is added to a base, the proton acceptor: H3O+ + OH‾ <==> H2O The solution is thus neutralized (pH=7). This can be measured with a pH meter, or it may be indicated with an indicator (not so exact).
What volume of 0.5M HNO3 is required to titrate 25 ml of a 0.2M NaOH solution to the point where the pH=7?
a. NaOH(aq) + HNO3(aq) ---> H2O + NaNO3(aq)
OHˉ + H3O+ <=> H2O
b. NaOH Molarity: 0.2mmol/mL
n = M*V = 0.2M * 25mL = 5 mmol
c. Reaction ratio NaOH : HNO3 is 1:1 so 5 mmoles of
HNO3(aq) will neutralize 25 mL of NaOH(aq)
d. V = n / M = 5mmol / 0.5mmol/mL= 10mL
I don't do this everyday so forgive me if my answer is not correct. This matter should be explained clearly in your chemistry literature. Most of my search results on the Internet seem to make things more complicated. Lots of cool theory but no solution or quick explanation, except my source beneath.
Cheers.
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