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In Chemistry / College | 2025-07-03

How many grams of potassium carbonate are needed to make 400.0 mL of 2.5 M solution?

$K_2 CO_3: 138.21 g / mol$

Asked by sawyerha1001

Answer (2)

To prepare 400.0 mL of a 2.5 M potassium carbonate solution, 138.21 grams of K₂CO₃ are required. First, convert the volume from mL to L, calculate the moles needed, and then find the mass using the molar mass provided. This calculation is essential for accurately creating solutions in chemistry.
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Answered by Anonymous | 2025-07-04

Convert the volume from mL to L: 400.0 m L = 0.4 L .
Calculate the number of moles of K 2 ​ C O 3 ​ : m o l es = 2.5 M × 0.4 L = 1.0 m o l .
Calculate the mass of K 2 ​ C O 3 ​ : ma ss = 1.0 m o l × 138.21 g / m o l = 138.21 g .
Therefore, the mass of potassium carbonate needed is 138.21 g ​ .

Explanation

Problem Analysis We are asked to find the mass of potassium carbonate ( K 2 ​ C O 3 ​ ) needed to make 400.0 mL of a 2.5 M solution. We are given the molar mass of K 2 ​ C O 3 ​ as 138.21 g/mol.

Convert Volume to Liters First, we need to convert the volume from milliliters (mL) to liters (L). To do this, we divide the volume in mL by 1000: V ( L ) = 1000 V ( m L ) ​ = 1000 400.0 ​ = 0.4 L

Calculate Moles of K2CO3 Next, we calculate the number of moles of K 2 ​ C O 3 ​ needed using the definition of molarity: M o l a r i t y = V o l u m e ( in L ) m o l es ​ So, m o l es = M o l a r i t y × V o l u m e ( in L ) = 2.5 L m o l ​ × 0.4 L = 1.0 m o l

Calculate Mass of K2CO3 Now, we calculate the mass of K 2 ​ C O 3 ​ needed using the formula: ma ss = m o l es × m o l a r ma ss = 1.0 m o l × 138.21 m o l g ​ = 138.21 g Therefore, 138.21 grams of potassium carbonate are needed.

Final Answer Therefore, to make 400.0 mL of a 2.5 M potassium carbonate solution, you need 138.21 grams of K 2 ​ C O 3 ​ .


Examples
In chemistry, preparing solutions of specific concentrations is a fundamental skill. For instance, in a lab setting, you might need to create a potassium carbonate solution to standardize an acid in a titration experiment. Knowing how to accurately calculate the mass of solute needed ensures the reliability and accuracy of your experimental results. This skill is also crucial in various industrial processes, such as manufacturing fertilizers or producing specialty chemicals, where precise concentrations are essential for product quality and process efficiency.

Answered by GinnyAnswer | 2025-07-04