Question

If 15.0 mL of 0.0250 M aqueous HNO3 is required to neutralize 10.0 mL of NaOH to the endpoint of the titration, what is the molarity of the NaOH?

Answer 1

The molarity of the NaOH solution is 0.0375 M. To determine the molarity of the NaOH solution we can apply the concept of stoichiometry in a titration reaction where the acid and base neutralize each other. Given that HNO3 and NaOH react in a 1:1 molar ratio:

Step-by-step explanation:

To determine the molarity of the NaOH solution, we can use the equation M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

In this case, we know that the initial molarity of HNO3 is 0.0250 M and the initial volume is 15.0 mL. The final volume, V2, is given as 10.0 mL. Plugging these values into the equation, we can solve for the final molarity, M2, of NaOH.

M1V1 = M2V2 => (0.0250 M)(15.0 mL) = M2(10.0 mL) => M2 = (0.0250 M)(15.0 mL)/(10.0 mL) = 0.0375 M

The molarity of the NaOH solution that neutralizes 15.0 mL of 0.0250 M HNO3 is calculated to be 0.0375 M, using the stoichiometry of the titration reaction and the volumes of the solutions used.

To determine the molarity of the NaOH solution we can apply the concept of stoichiometry in a titration reaction where the acid and base neutralize each other. Given that HNO3 and NaOH react in a 1:1 molar ratio:

1. Calculate the number of moles of HNO3 used in the reaction.
   
2. Since the stoichiometry of the reaction is 1:1, 0.000375 moles of NaOH are required for neutralization.
3. Calculate the molarity of the NaOH solution.
   

Therefore, the molarity of the NaOH solution is 0.0375 M.