Question

0.22 L of HNO3 is titrated to equivalence using 0.10 L of 0.1 MNaOH. What is the concentration of the HNO3?

Express the concentration of HNO3 in molarity.

Answer 1

The concentration of HNO3 is calculated to be 0.045 M by using dimensional analysis to find the moles of NaOH and HNO3, which react in a 1:1 ratio, and then dividing by the volume of HNO3 used.

Step-by-step explanation:

To find the concentration of HNO3, first we need to calculate the number of moles of NaOH used in the titration. Since the NaOH solution has a concentration of 0.1 M, we can use dimensional analysis to find the moles of NaOH:
Moles of NaOH = Volume of NaOH (L) × Concentration of NaOH (M)
Moles of NaOH = 0.10 L × 0.1 M
The reaction between HNO3 and NaOH is as follows:
HNO3 + NaOH → NaNO3 + H2O
Since both HNO3 and NaOH react in a 1:1 molar ratio, the moles of NaOH are equal to the moles of HNO3 at the equivalence point. Therefore, the moles of HNO3 is also 0.01 mol.
Now, we can calculate the molarity of HNO3 by dividing these moles by the volume of the HNO3 solution in liters.
Concentration of HNO3 (M) = Moles of HNO3 / Volume of HNO3 (L)
Concentration of HNO3 (M) = 0.01 mol / 0.22 L

Concentration of HNO3 = 0.045 M

Answer 2

Answer:- Molarity of the acid solution is 0.045M.

Solution:- The balanced equation for the reaction of given acid and base is:

`HNO_3+NaOH\rightarrow NaNO_3+H_2O`

From the balanced equation, they react in 1:1 mol ratio. So, we could easily solve the problem using the equation:

`M_aV_a=M_bV_b`

where,
`M_a` is the molarity of acid,
`M_b` is the molarity of base,
`V_a` is the volume of acid and
`V_b` is the volume of base.

Let's plug in the given values in the equation:

`M_a(0.22L)=0.1M(0.10L)`

on rearranging the above equation:

`M_a=(0.1M(0.10L))/(0.22L)`

`M_a` = 0.045M

So, the molarity of the acid solution is 0.045M.