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

Answer:

$M_{acid}=0.075M$

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:

$M_{acid}=\frac{M_{base}V_{base}}{V_{acid}}$

Therefore, by plugging in we obtain:

$M_{acid}=\frac{0.12L*0.2M}{0.32L}\\\\ M_{acid}=0.075M$

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