Calculate the number of milliliters of 0.738 M NaOH required to precipitate all of the Mg2+ ions in 148 mL of 0.541 M MgSO4 solution as Mg(OH)2. The equation for the reaction is:

MgSO4(aq) + 2 NaOH(aq) Mg(OH)2(s) + Na2SO4(aq)

_____________mL NaOH

How many mols MgSO4 so you have? That is moles = M x L = 0.541 x 0.148 = ?

Using the cofficients in the balanced equation, convert moles MgSO4 to moles NaOH. That is
?moles MgSO4 x (2 moles NaOH/1 mol MgSO4) = ?moles MgSO4 x 2/1 = ? moles MgSO4.

Then M = moles/L soln
MNaOH = moles NaoH/L NaOH
You know M NaOH and moles NaOH, solve for L NaOH and convert to mL.

To calculate the number of milliliters of 0.738 M NaOH required, we first need to determine the stoichiometry of the reaction between MgSO4 and NaOH.

From the balanced chemical equation:
MgSO4(aq) + 2 NaOH(aq) → Mg(OH)2(s) + Na2SO4(aq)

We can see that for every 1 mole of MgSO4, 2 moles of NaOH are required. This means that the molar ratio of MgSO4 to NaOH is 1:2.

Now let's calculate the moles of MgSO4 present in the solution:
Moles of MgSO4 = volume (in liters) * molarity
= 148 mL * (1 L / 1000 mL) * 0.541 mol/L
= 0.0801088 mol

Since the molar ratio of MgSO4 to NaOH is 1:2, the moles of NaOH required is twice the moles of MgSO4:
Moles of NaOH = 2 * 0.0801088 mol
= 0.160216 mol

Now we can convert the moles of NaOH to volume using its molarity:
Volume (in liters) = Moles / Molarity
= 0.160216 mol / 0.738 mol/L
= 0.217197 L

Finally, we can convert the volume from liters to milliliters:
Volume (in mL) = 0.217197 L * (1000 mL / 1 L)
= 217.197 mL

Therefore, the number of milliliters of 0.738 M NaOH required to precipitate all of the Mg2+ ions in the solution is approximately 217.197 mL.

To calculate the number of milliliters of NaOH required, we need to use the balanced chemical equation and stoichiometry.

First, let's determine the moles of Mg2+ ions present in the 148 mL of 0.541 M MgSO4 solution. We can use the formula:

moles = concentration (in M) x volume (in liters)

Moles of MgSO4 = 0.541 M x (148 mL / 1000 mL/ L) = 0.079 moles

According to the balanced chemical equation, 1 mole of MgSO4 reacts with 2 moles of NaOH to form 1 mole of Mg(OH)2. This means that the ratio of moles of NaOH to moles of MgSO4 is 2:1.

Therefore, the moles of NaOH required to react with all the moles of MgSO4 can be calculated using the stoichiometric ratio:

moles of NaOH = 2 x (moles of MgSO4) = 2 x 0.079 moles = 0.158 moles

Now, let's determine the volume of 0.738 M NaOH solution required to have 0.158 moles of NaOH. We can rearrange the formula to solve for the volume:

volume (in L) = moles / concentration (in M)

Volume of NaOH = 0.158 moles / 0.738 M = 0.214 L

Finally, convert the volume in liters to milliliters:

Volume of NaOH = 0.214 L x 1000 mL/L = 214 mL

Therefore, 214 mL of 0.738 M NaOH solution is required to precipitate all the Mg2+ ions in the 148 mL of 0.541 M MgSO4 solution as Mg(OH)2.