Question

4.60 mL of 0.1852 M HNO3 is titrated to the phenolphthalein indicator endpoint with 27.35 mL of a KOH solution. What is the molarity of the KOH solution

Answer 1

Answer:

$M_{base}=0.0311M$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve for the molarity of the KOH solution by knowing this base react in a 1:1 mole ratio with nitric acid, HNO3; thus, we can write the following equation, as their moles are the same at the endpoint:

$n_{acid}=n_{base}$

Which in terms of molarities and volumes is:

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

Thus, we solve for the molarity of the base (KOH) to obtain:

$M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} \\\\M_{base}=\frac{4.60mL*0.1852M}{27.35mL}\\\\M_{base}=0.0311M$

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