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Question Video: Using Strong Acid-Strong Base Titration Data to Calculate the Concentration of the Base Chemistry

The table below shows the results of titrating 0.02 dm³ of Ba(OH)₂ against 0.2 M of hydrochloric acid. Using the results in the table and discounting any anomalies, determine the concentration of Ba(OH)₂. Give your answer to 2 decimal places.

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Video Transcript

The table below shows the results of titrating 0.02 decimeters cubed of barium hydroxide against 0.2 molar of hydrochloric acid. Using the results in the table and discounting any anomalies, determine the concentration of barium hydroxide. Give your answer to two decimal places.

Looking at the data, we can see that titration one and titration three required very similar volumes of hydrochloric acid. Titration two, however, required over 10 milliliters more of hydrochloric acid. This data is inconsistent with the other two trials and should be discounted as an anomaly. We can then take the volume of HCL solution added in titration one and titration three and find the average volume of HCL. Performing the calculation gives us an average volume of 61.5 milliliters.

Next, we can write a balanced chemical equation between barium hydroxide and hydrochloric acid. We can see that the reaction of barium hydroxide, a base, with hydrochloric acid, the acid, produces barium chloride — a salt — and water. Be careful and ensure that the chemical equation is balanced by adding appropriate coefficients before continuing on with the rest of the problem.

Next, recall the 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 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. This titration required 0.02 decimeters cubed of barium hydroxide. Recall that one decimeter cubed is the same as one liter. We can then add our volume to the table with the unit liters instead of decimeters cubed. The barium hydroxide solution was titrated against a 0.2-molar solution of hydrochloric acid. We can add this concentration to our table in the appropriate box, recognizing that molar and moles per liter are equivalent units.

We also know the average volume of hydrochloric acid used in the experiment. The volume is in milliliters but must be converted into liters in order for the liters and the concentration unit to cancel when solving. Recognize that 1000 milliliters are equivalent to one liter. We can multiply the 61.5 milliliters by one liter per 1000 milliliters. The unit milliliters will cancel, giving us a volume of 0.0615 liters. Ultimately, we would like to solve for the concentration of barium hydroxide.

Now that the given values have been filled into the table, we are ready to solve the problem. We can substitute our hydrochloric acid concentration and volume into the key equation to determine the number of moles of hydrochloric acid to be 0.0123 moles. Now that we know the number of moles of acid used, we can determine the number of moles of base used in the titration.

Looking at our balanced chemical equation, we can see that the molar ratio of barium hydroxide to hydrochloric acid is one to two. As the molar ratio is not one to one, we will need to perform a calculation to convert moles of hydrochloric acid into moles of barium hydroxide. We begin this process with the moles of hydrochloric acid. We then multiply this value by the molar ratio written as a fraction with moles of hydrochloric acid in the denominator so that the units cancel out. We perform the calculation and determine the number of moles of barium hydroxide to be 0.00615 moles.

Next, we can rearrange our key equation to solve for the concentration of barium hydroxide. We can substitute our barium hydroxide amount and volume and determine the concentration of barium hydroxide to be 0.3075 molar. But the question asks us for an answer rounded to two decimal places. Rounding our answer appropriately gives us a final concentration of 0.31 molar.

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