How many mL of a 0.5M Ca(OH)2 solution are needed to neutralize 38 mL of a 2M HNO3 solution?
ANSWER IS: x = multiply / = divide
38mL HNO3 X 1 L HNO3 x 2M HNO3 x 1 L Ca(OH)2 x 2M Ca(OH)2 x 1000 mL / 1000 mL/1L HNO3/..5mL Ca(OH)2/1M HNO3/1L = 304 mL Ca(OH)2
ANSWER IS: x = multiply / = divide
38mL HNO3 X 1 L HNO3 x 2M HNO3 x 1 L Ca(OH)2 x 2M Ca(OH)2 x 1000 mL / 1000 mL/1L HNO3/..5mL Ca(OH)2/1M HNO3/1L = 304 mL Ca(OH)2
To find out how many mL of 0.5M Ca(OH)2 solution are needed to neutralize 38 mL of 2M HNO3 solution, you can use the concept of molar ratios.
Before we start, let's break down the given information:
- 38 mL of 2M HNO3 solution
- 0.5M Ca(OH)2 solution
Now, let's set up the equation using the information given and the concept of molar ratios:
38 mL HNO3 x (1 L HNO3 / 1000 mL HNO3) x (2 mol HNO3 / 1 L HNO3) x (1 mol Ca(OH)2 / 2 mol HNO3) x (1000 mL Ca(OH)2 / 0.5 mol Ca(OH)2)
Let's break down the equation step by step to understand the calculations:
- First, we convert 38 mL of HNO3 to L by multiplying by the conversion factor (1 L HNO3 / 1000 mL HNO3). This gives us the value in liters to match the molarity unit.
- Next, we use the molarity of HNO3 (2 mol HNO3 / 1 L HNO3) to convert from HNO3 to moles of HNO3. The unit "L HNO3" cancels out, leaving us with moles.
- Then, we use the balanced chemical equation to determine the mole ratio between HNO3 and Ca(OH)2. According to the equation, 2 moles of HNO3 react with 1 mole of Ca(OH)2.
- After that, we use the molar ratio of Ca(OH)2 (1 mol Ca(OH)2 / 2 mol HNO3) to convert from moles of HNO3 to moles of Ca(OH)2. The "mol HNO3" units cancel out, leaving us with moles of Ca(OH)2.
- Finally, we convert from moles of Ca(OH)2 to mL of Ca(OH)2 using the molarity of Ca(OH)2 (1000 mL Ca(OH)2 / 0.5 mol Ca(OH)2). The unit "mol Ca(OH)2" cancels out, leaving us with mL of Ca(OH)2.
By multiplying all the values together and canceling units, we get:
38 mL HNO3 x (1 L HNO3 / 1000 mL HNO3) x (2 mol HNO3 / 1 L HNO3) x (1 mol Ca(OH)2 / 2 mol HNO3) x (1000 mL Ca(OH)2 / 0.5 mol Ca(OH)2) = 304 mL Ca(OH)2
Therefore, 304 mL of 0.5M Ca(OH)2 solution is needed to neutralize 38 mL of 2M HNO3 solution.