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A 25.00 mL sample of sodium hydroxide was titrated (neutralized) with 36.4 mL of 0.125 M sulfuric acid. What is the concentration of the sodium hydroxide solution?Write your answer to three digits - for example 0.468 and leave of the unit M. Round off your answer to three digits - so 0.4682 is written as 0.468.

User Angeli
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2 Answers

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To find the concentration of the sodium hydroxide solution, we need to use the equation:

moles of acid = moles of base

We can start by calculating the moles of acid. We know that the volume of the sulfuric acid is 36.4 mL and the concentration is 0.125 M. We can use the formula:

moles of acid = concentration x volume (in liters)

First, we need to convert the volume to liters:

36.4 mL = 0.0364 L

Now we can calculate the moles of acid:

moles of acid = 0.125 M x 0.0364 L = 0.00455 moles

Since the acid and the base react in a 1:2 ratio, we know that there must be twice as many moles of sodium hydroxide as there are moles of sulfuric acid. Therefore, the moles of sodium hydroxide can be calculated as:

moles of base = 2 x moles of acid = 2 x 0.00455 moles = 0.00910 moles

We also know the volume of the sodium hydroxide solution is 25.00 mL, which is 0.02500 L. Now we can use the equation:

concentration of base = moles of base / volume of base

to calculate the concentration of the sodium hydroxide solution:

concentration of base = 0.00910 moles / 0.02500 L = 0.364 M

Rounding off to three digits, we get:

concentration of base = 0.364
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ANSWER: Answer is in image

A 25.00 mL sample of sodium hydroxide was titrated (neutralized) with 36.4 mL of 0.125 M-example-1
User Rimil Dey
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