Question

0.32 L of HNO3 is titrated to equivalence using 0.12 L of 0.2 M NqOH. What is the concentration of the HNO3

Answer 1

`M_(acid)=0.075M`

Step-by-step explanation:

Hello!

In this case, since the reaction between NaOH and HNO3 is:

`NaOH+HNO_3\rightarrow NaNO_3+H_2O`

Whereas there is a 1:1 mole ratio between the acid and base, thus, we can write:

`M_(acid)V_(acid)=M_(base)V_(base)`

In such a way, solving for the concentration of the acid, HNO3, we obtain:

`M_(acid)=(M_(base)V_(base))/(V_(acid))`

Therefore, by plugging in we obtain:

`M_(acid)=(0.12L*0.2M)/(0.32L)\\\\ M_(acid)=0.075M`

Best regards!