Knowing the milk of magnesia contains 8 g% of Mg (OH) 2, we used 1 g milk of magnesia and added 50 mL of 0.100 M HCl, determine the volumeainda of a solution of 0.105 M NaOH to titrate the excess HCl 0.100 M
Given 1gm of MM contains 8% Mg(OH)2 (f.wt. = 58.33 g/mol)
Gms Mg(OH)2
= 8% of 1gm MM = 0.08gms Mg(OH)2
= 0.00137mol Mg(OH)2
Mg(OH)2 in solution => 2moles OH^-
Moles OH^- = 2(0.00137)mole OH^-
= 0.00274mole OH^-
Adding 50ml of 0.100M HCl
=> 0.005mole HCl + 0.00274 mole OH^-
=>(0.0050 - 0.00274)mole excess(H^+)
=> 0.0023mole excess (H^+)
=> (0.0023/0.050)M H^+
=> 0.045M in excess H^+
Titrating 50ml of 0.045M H^+ with 0.105M NaOH
=>(M x V)acid = (M x V)base
=>(0.045M)(50ml)=(0.105M)(Vol Base)
=> Vol Base = (0.045M)(50ml)/(0.105M)
=> Vol Base = 21.4ml of 0.105M NaOH
To determine the volume of a solution of 0.105 M NaOH required to titrate the excess HCl, we need to first calculate the moles of HCl present in the 50 mL of 0.100 M HCl solution.
Moles of HCl can be calculated using the formula:
moles = concentration (M) × volume (L)
Given that the concentration of HCl is 0.100 M and the volume is 50 mL (or 0.050 L), we can calculate the moles of HCl:
moles HCl = 0.100 M × 0.050 L = 0.005 moles
Next, we need to determine the moles of Mg(OH)2 in 1 g of milk of magnesia. Since we know that the milk of magnesia contains 8 g% of Mg(OH)2, we can calculate the moles using the molar mass.
The molar mass of Mg(OH)2 is:
Mg: 24.31 g/mol
O: 16.00 g/mol (2 atoms)
H: 1.01 g/mol (2 atoms)
Summing up, we have:
Molar mass Mg(OH)2 = 24.31 g/mol + 16.00 g/mol × 2 + 1.01 g/mol × 2 = 58.33 g/mol
Now we can calculate the moles of Mg(OH)2:
moles Mg(OH)2 = mass / molar mass = 1 g / 58.33 g/mol ≈ 0.017 moles
Since the reaction between HCl and Mg(OH)2 is in a 1:2 ratio, we can determine that the moles of HCl required to react with the moles of Mg(OH)2 are half the amount of moles of Mg(OH)2. Hence, the moles of HCl in excess are:
moles HCl in excess = moles HCl - (0.5 × moles Mg(OH)2)
= 0.005 moles - (0.5 × 0.017 moles)
= 0.005 moles - 0.0085 moles
= 0.0045 moles
Now, we need to determine the moles of NaOH required to react with the excess HCl. Since the reaction between HCl and NaOH is in a 1:1 ratio, the moles of NaOH required are the same as the moles of HCl in excess, which is 0.0045 moles.
Finally, to find the volume of the 0.105 M NaOH solution required, we can use the formula:
volume (L) = moles / concentration (M)
volume (L) = 0.0045 moles / 0.105 M
≈ 0.0429 L
Therefore, the volume of the 0.105 M NaOH solution required to titrate the excess HCl is approximately 42.9 mL.