Question

1. A flask contains 4.00 g of solid sodium hydroxide. Some water is added to the flask to dissolve the sodium hydroxide, and enough water is added to fill the flask to a volume of 50.0 mL. What volume of 1.00 M HCl is needed to neutralize the sodium hydroxide?

A. 50.0 mL

B. 100.0 mL

C. 200.0 mL

2. A beaker contains 100.0 mL of a 0.100 M solution of sulfuric acid. When 100.0 mL water is added to this solution, what is the concentration of the diluted sulfuric acid?

A. 0.0250 M

B. 0.0500 M

C. 0.200 M

3. How many moles of NaOH are required to react completely with 1.00 mole of nitric acid?

A. 1.00 mol

B. 0.500 mol

C. 2.000 mol

Please explain. Thank you!

Answer 1

Sorry I do not know how to explain this.
However I found that @Kevin, is great at this stuff.
He will be online soon, so stay online.

yw

Answer 2

@Kevin, make sure you explain it to this guy.

Answer 3

I could do this, but it's pretty late in my part of the world and with my schedule, so I can get on this first thing in the morning if no one else answers, seem ok @KJ4UTS?

Answer 4

@SapphireMoon no problem, I appreciate any help you can give me. This is a new unit for me and im having some difficulty understanding it. Thanks again :)

Answer 5

Answer 1 :-
concentration of NaOH after dissolution of 50 ml of water in it =
\[\text{Molarity }= \frac{\text{no. of moles of solute} }{\text{volume of solution in liters}}\]
number of moles of NaOH= \(\Large\frac{4}{40}\)
volume of solution in liters = \(\Large \frac{50}{1000}\) L

concentration of NaOH = \(\Large \frac{4/40}{50/1000}\) M

Remember the fact that during Neutralization reaction

Milli-equivalents of acid = Milliequivalents of base

(number of milliequivalents = Molarity * n-factor * volume of solution in milliliters)

Answer 6

n-factor for HCl =1
n-factor for NaOH =1

Let the volume of HCl required for neutralization be 'V' ml

Milliequivalents of HCl = Milliequivalents of NaOH

1M * 1 * V= \(\Large \frac{4/40}{50/1000}\) * 1 * 50 .......(1)

solve equation (1) yo get 'V'.

remember u will gt volume in milliliters

Answer 7

Answer :- 2
During dilution number of moles of the solute do not alter so

\[M_1 V_1=M_2 V_2\]
\(M_1\) = initial concentration
\(M_2\)=final concentration
\(V_1\)=Initial volume of solution
\(V_2\)=final volume of solution

can u apply this rule now to find answer for question 2 ?

Answer 8

Answer :-
As i told u earlier that during neutralization reaction

(Milli-equivalents of acid = Milliequivalents of base)

(milliequivalent of a compound = no. of moles of that substance * n-factor)

milliequivalents of nitric acid = 1*1=1
(as both no. of moles & n-factor of \(HNO_3\) are 1)

Let the number of moles of NaOH required for neutralization of \(HNO_3\) be 'n'.

milliequivalents of NaOH = n*1 (as no. of moles of NaOH is n & n-factor is 1)

Therefore,
milliequivalents of nitric acid=milliequivalents of NaOH
1*1=n*1

Answer 9

1)
M = gr/Mr x 1000/p
M = 4/40 x 1000/50
M = 2

M1 x V1 = M2 x V2
2 x 50 = 1 x V2
V2 = 100

2)
M1 x V2 = M2 x V2
(just like the question on number 1)

3)
HNO3 + NaOH -> NaNO3 + H2O

HNO3 -> H^+ + NO3^- -> valency of H+ = 1
NaOH -> Na^+ + OH^- -> valency of OH^- = 1

M1 x V1 x valency = M2 x V2 x valency
mole1 x valency = mole2 x valency
1 x 1 = mole2 x 1