Chem help?

Question

What is the molarity of a nitritic acid solution, HNO3, if 20.0 mL of the solution is needed to exactly neutralize 10.0 mL of a 1.67 M NaOH solution?

a- 3.34 M
b- 1.67 M
c- 0.334 M
d- 0.835 M

thanks in advance =)

Answer 1

NaOH + HNO3 --> NaNO3 + H2O

=1.67molNaOH/L * 0.01L * 1molHNO3/1mol NaOH * 1/0.02L
=0.835M HNO3 [Answer: D]

You weren't kidding about your number of questions...

Answer 2

first, write the balanced reaction equation;

HNO3 + NaOH --> NaNO3 + H2O

Then, write down the values you are given;

c = concentration/molarity (molL^-1 = M)
V = volume (L)
n = moles

V(HNO3) = (20/1000)L = 0.02L

c(NaOH) = 1.67M
V(NaOH) = (10/1000)L = 0.01L

n(NaOH) = c(NaOH) x V(NaOH)
n(NaOH) = 1.67M x 0.01L
n(NaOH) = 0.0167 mol

According to the balanced reaction equation, the mole ratio between NaOH and HNO3 is 1:1

thus n(HNO3) = n(NaOH) = 0.0167 mol

Thus;

c(HNO3) = n(HNO3) / V(HNO3)
c(HNO3) = 0.0167 mol / 0.02L
c(HNO3) = 0.835 M

Thus the answer is 0.835 M of HNO3

Hope this helps:-)

Answer 3

n = CV

NaOH solution:
10ml = 0.01 L
n=1.67 * 0.01
n=0.0167

theres a 1:1 ratio in formula so
HNO3:
20ml = 0.02L
0.0167 = C * 0.02
C = 0.0167/0.02
C = 0.835

answer = (d) 0.835M