What volume of 3.0M sodium hydroxide will be needed to neutralize 2.0 L of 1.0M hydrochloric acid?
How many mols HCl? M x L = mols.
HCl + NaOH ==> NaCl + H2O
1:1; therefore, mols NaOH = moles HCl.
M NaOH = moles NaOH/L NaOH. Solve for L.
Naoh+Hcl==>Nacl+H2o
1:1for mole
Mole of Naoh = mole of Hcl
That is mean :
M of Naoh ×volume of Naoh = M of Hcl ×volume of Hcl
3×v=1×2 ====»3v =2
V=2\3 liter
The volume of Naoh=2/3 liter
To find the volume of a solution needed to neutralize another solution, we can use the concept of molar ratios.
First, let's write a balanced chemical equation for the neutralization reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl):
NaOH + HCl -> NaCl + H2O
From the equation, we can see that one mole of NaOH reacts with one mole of HCl.
Next, we need to determine the number of moles of HCl in 2.0 L of a 1.0 M solution. Remember that the molarity (M) of a solution is defined as moles of solute per liter of solution. Therefore, the number of moles of HCl can be calculated as follows:
Number of moles of HCl = Molarity of HCl x Volume of HCl
Number of moles of HCl = 1.0 mol/L x 2.0 L = 2.0 moles
Since the molar ratio between NaOH and HCl is 1:1, we need an equal number of moles of NaOH to neutralize the HCl.
Finally, we can determine the volume of 3.0 M NaOH solution needed to react with 2.0 moles of HCl:
Volume of NaOH = Number of moles of NaOH / Molarity of NaOH
Volume of NaOH = 2.0 moles / 3.0 mol/L = 0.67 L
Therefore, approximately 0.67 liters (or 670 mL) of 3.0 M NaOH solution will be needed to neutralize 2.0 liters of 1.0 M HCl.