Asked by Bekah

How many milliliters of 0.100 M HNO3 are needed to neutralize the following solutions?
A. 45.0 mL of 0.667 M KOH
B. 58.5 mL of 0.0100 M Al(OH)3
Answered by DrBob222
Writ the equations and balance them.
HNO3 + KOH ==> KNO3 + H2O
3HNO3 + Al(OH)3 ==> 3H2O + Al(NO3)3

millimoles KOH = mL x M = 45.0*0.667 = 30.015
Convert mmoles KOH to mmoles HNO3 using the coefficients in the balanced equation. That is done as
mmoles HNO3 = mmoles KOH x (1 mole HNO3/1 mole KOH) = 30.015 x (1/1) = 30.015 mmoles HNO3.
Then M = mmoles/mL and solve for mL = mmoles/M = 30.015/0.100 = 300.15 mL. Round to the correct number of sig figures.
The second part is done the same way but the ratio of acid/base is not 1:1 as it is with KOH/HNO3.
Answered by sharad
When Aluminium is added to KOH then solution is?
Answered by Anonymous
7676
Answered by Jess
300.15 mL
Answered by Jeel
B. 17.55 ml
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