Knowing that the milk of magnesia contains 8 g% of Mg (OH)2, using 1 g of milk of magnesia and added 50 mL of 0.100 M HCl, determine the required volume of a solution of 0.105 M NaOH to titrate the excess HCl 0.100 M.
20 mL of a solution of HCl concentration, neutralized 0,50 g of Na2CO3. What is the Molarity of the acid solution?
To solve the problem, we need to calculate the volume of the NaOH solution required to titrate the excess HCl. Here are the steps:
Step 1: Calculate the moles of Mg(OH)2 in 1 g of milk of magnesia.
- The milk of magnesia contains 8 g% of Mg(OH)2. This means that in 100 g of milk of magnesia, there are 8 g of Mg(OH)2.
- Using this information, we can calculate the moles of Mg(OH)2 in 1 g of milk of magnesia.
- The molar mass of Mg(OH)2 is 58.32 g/mol.
- So, moles of Mg(OH)2 = (1 g / 100 g) * (1 mol / 58.32 g) = 0.0171 mol
Step 2: Calculate the moles of HCl used.
- We have added 50 mL of 0.100 M HCl. To calculate the moles of HCl used, we need to use the formula: moles = concentration * volume (in liters).
- Volume of HCl used = 50 mL = 50/1000 L = 0.05 L
- Moles of HCl used = 0.100 M * 0.05 L = 0.005 mol
Step 3: Calculate the moles of NaOH required to react with the excess HCl.
- The balanced chemical equation for the reaction between HCl and NaOH is: HCl + NaOH -> H2O + NaCl
- From the equation, we see that 1 mole of HCl reacts with 1 mole of NaOH.
- So, the moles of NaOH required to react with the excess HCl is also 0.005 mol.
Step 4: Calculate the volume of the NaOH solution required.
- We have the concentration of NaOH solution, which is 0.105 M.
- Using the formula: moles = concentration * volume (in liters), we can rearrange to solve for volume.
- Volume of NaOH solution required = moles / concentration = 0.005 mol / 0.105 M = 0.0476 L
Step 5: Convert the volume to milliliters (mL).
- Volume of NaOH solution required = 0.0476 L * 1000 mL/L = 47.6 mL
Therefore, the required volume of a solution of 0.105 M NaOH to titrate the excess 0.100 M HCl is approximately 47.6 mL.