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.
1-g of MM contains 8 wt% Mg(OH)2 (f.wt=58.33g/mole)
8% of 1gm = 0.08(1)g Mg(OH)2
= 0.08g Mg(OH)2
= (0.08/58.33)mole Mg(OH)2
= 0.00137 mole Mg(OH)2
Mg(OH)2 in solution => Mg^+2 + 2OH^-
moles of OH- = 2(0.00137)mole OH^-
= 0.00274mole OH^
Adding 50ml of 0.100M HCl
= 0.050(0.10)mole HCl
= 0.005mole HCl
0.005mole HCl + 0.00274mole OH^-
=> (0.005 - 0.00274)mole excess HCl
=> 0.0023mole excess HCl
=> 0.0023mole excess H^+
Concentration of excess (H^+) in 50ml solution:
[H^+] = (0.0023/0.050)M = 0.045M
Titrating 50ml of 0.045M H^+ with 0.105M NaOH
(M x V)acid = (M x V)base
(0.045M)(50ml)=(0.105M)(Vol of Base)
Vol of Base = (0.045M)(50ml)/(0.105M)
Vol of 0.105M NaOH needed = 21.4ml
Thanks :)
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To determine the required volume of the 0.105 M NaOH solution needed to titrate the excess HCl, we can use a balanced chemical equation and the concept of stoichiometry.
Here's the balanced equation for the reaction between HCl and NaOH:
HCl + NaOH → H2O + NaCl
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. So, we need to find the number of moles of HCl first.
Given that you added 50 mL of 0.100 M HCl solution, we can calculate the moles of HCl using the formula:
moles = concentration (M) x volume (L)
First, convert the volume from mL to L by dividing by 1000:
50 mL ÷ 1000 = 0.05 L
Now, calculate the moles of HCl:
moles of HCl = 0.100 M x 0.05 L = 0.005 moles
Since the balanced equation shows a 1:1 mole ratio between HCl and NaOH, we need an equal number of moles of NaOH to completely react with the HCl.
Now, let's determine the amount of HCl that remains after reacting with the 1 g of milk of magnesia. Since the milk of magnesia contains 8 g% of Mg(OH)2, this means that 1 g of milk of magnesia contains 0.08 g of Mg(OH)2.
To convert the mass of Mg(OH)2 to moles, we need to divide by its molar mass. The molar mass of Mg(OH)2 is:
Mg: 24.31 g/mol
O: 16.00 g/mol (x2)
H: 1.01 g/mol (x2)
Molar mass of Mg(OH)2 = 24.31 + 2(16.00) + 2(1.01) = 58.33 g/mol
Now, calculate the number of moles of Mg(OH)2 in 1 g of milk of magnesia:
moles of Mg(OH)2 = mass (g) / molar mass (g/mol) = 0.08 g / 58.33 g/mol ≈ 0.00137 moles
Since 1 mole of Mg(OH)2 reacts with 2 moles of HCl, we can calculate the number of moles of HCl that reacted with 1 g of milk of magnesia:
moles of HCl reacted = 2 x moles of Mg(OH)2 = 2 x 0.00137 moles = 0.00274 moles
To find the number of moles of HCl that remain:
moles of HCl remaining = moles of HCl initially added - moles of HCl reacted
= 0.005 moles - 0.00274 moles
= 0.00226 moles HCl
Finally, using the concentration (0.105 M) and the number of moles of HCl remaining (0.00226 moles), we can determine the volume of 0.105 M NaOH solution required to titrate the excess HCl:
volume (L) = moles / concentration (M)
= 0.00226 moles / 0.105 M
≈ 0.0215 L
Convert the volume from liters to milliliters:
0.0215 L x 1000 = 21.5 mL
Therefore, the required volume of a 0.105 M NaOH solution to titrate the excess HCl is approximately 21.5 mL.