# Question Video: Calculating the Concentration of Lithium Hydroxide in Millimolars Via Titration with Sulfuric Acid Chemistry

A standard solution of 0.25 M H₂SO₄ is used to determine the concentration of a 220 mL LiOH solution. The addition of 143 mL of H₂SO₄ resulted in complete neutralization. What is the concentration of the LiOH solution? Give your answer in millimolars.

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### Video Transcript

A standard solution of 0.25 molar H2SO4 is used to determine the concentration of a 220-milliliter LiOH solution. The addition of 143 milliliters of H2SO4 resulted in complete neutralization. What is the concentration of the LiOH solution? Give your answer in millimolars.

Recall that in a neutralization reaction, an acid reacts with a base to produce a salt and water. Let’s take a look at the question and identify each component of the reaction. H2SO4, sulfuric acid, is the acid. And LiOH, lithium hydroxide, is the base. They react to produce water, chemical formula H2O, and a salt. The water was produced when the hydrogen from the acid combined with the hydroxide from the base. This means that the salt must be produced from the combination of the sulfate from the acid and the lithium from the base.

Notice in the sulfuric acid formula that there are two hydrogen ions for every sulfate ion. Hydrogen and lithium are found in the same group on the periodic table and will bond similarly. This means that when we wrote the formula for the salt, we should have written two lithium ions for every sulfate ion. We have identified all of the species in this reaction. However, we still need to make sure that the equation is balanced. We will need to place a coefficient of two in front of the lithium hydroxide and in front of the water. The equation is now balanced and we can continue on with the question.

A standard solution — that is, a solution with a known concentration — was used to determine the concentration of another solution. This is most often carried out via a titration experiment. As such, let’s recall our key equation for solving titration problems. 𝑛 equals 𝑐𝑣, where 𝑛 represents the amount in moles, 𝑐 is the concentration in moles per liter, and 𝑣 is the volume in liters. We can make a table to help us match the values given in the question with the variables of our key equation. We will also record the molar ratio of the acid and base.

A standard solution of 0.25 molar sulfuric acid was used. This is the concentration of the sulfuric acid. Recognize that molar and moles per liter are equivalent units. The sulfuric acid solution was combined with 220 milliliters of lithium hydroxide solution. The volume is given in milliliters but must be converted into liters in order for the liters in the concentration unit to cancel when solving. Recall that 1000 milliliters are equal to one liter. To convert the volume into liters, we simply take the 220 milliliters and multiply it by one liter per 1000 milliliters. This gives us a volume of 0.22 liters. The volume of sulfuric acid used in the experiment was 143 milliliters.

Once again, we will need to convert by multiplying the 143 milliliters by one liter per 1000 milliliters. This gives us a volume of 0.143 liters. The question is asking us to determine the concentration of the lithium hydroxide. We can place a question mark in the appropriate box in the table. Now that the given values have been filled into the table, we are ready to solve the problem. We can substitute the sulfuric acid concentration and volume into the key equation to determine the number of moles of sulfuric acid to be 0.03575 moles.

Now that we know the number of moles of acid used, we can determine the number of moles of base used. Looking at the coefficients in the balanced chemical equation, we can see that the molar ratio of sulfuric acid to lithium hydroxide is one to two. As the molar ratio is not one to one, we will need to perform a calculation in order to convert the moles of sulfuric acid into moles of lithium hydroxide. We begin the calculation with the moles of sulfuric acid. We then multiply by the molar ratio written as a fraction, with moles of sulfuric acid in the denominator so that the units cancel out. We perform the calculation and determine the number of moles of lithium hydroxide to be 0.0715 moles.

Next, we can rearrange the key equation to solve for the concentration of lithium hydroxide. We can substitute our lithium hydroxide amount and volume and determine the concentration of lithium hydroxide to be 0.325 molar. However, the question asks us to give our answer in millimolars. 1000 millimolar is equal to one molar. We can take our concentration in molar and multiply it by 1000 millimolar per one molar. This gives us a final lithium hydroxide concentration of 325 millimolar.