How many ml of a .25M HCL would be required to neutralize completely 2.5g of NaOH
Write the equation and balance it.
NaOH + HCl ==> NaCl + H2O
mols NaOH = 2.5g/molar mass NaOH.
mols HCl required = mols NaOH (from the equation).
Then mol HCl = M(HCl) x L(HCl). You have the molarity of the HCl and you have the mols of HCl from your calculation, which leaves just one unknown (the L). Convert that to mL.
To find out how many milliliters of a 0.25M HCl solution would be required to neutralize completely 2.5g of NaOH, we need to use the concept of molarity and the balanced chemical equation of the reaction between HCl and NaOH.
The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water.
To calculate the number of moles of NaOH in 2.5g, we need to divide the given mass by the molar mass of NaOH. The molar mass of NaOH is the sum of the atomic masses of Na (22.99 g/mol) + O (16.00 g/mol) + H (1.01 g/mol), which equals 39.99 g/mol.
Number of moles of NaOH = mass of NaOH / molar mass of NaOH
Number of moles of NaOH = 2.5g / 39.99 g/mol ≈ 0.0625 mol NaOH
According to the balanced equation, 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the number of moles of HCl required to completely neutralize the given amount of NaOH is also 0.0625 mol.
Now, we can calculate the volume of the 0.25M HCl solution needed. The term "0.25M" implies that there are 0.25 moles of HCl in 1 liter of HCl solution.
Volume of HCl solution (in liters) = number of moles of HCl / molarity of HCl
Volume of HCl solution = 0.0625 mol / 0.25 mol/L = 0.25 L = 250 mL
Therefore, 250 milliliters (mL) of a 0.25M HCl solution would be required to neutralize completely 2.5g of NaOH.