How many mL of 0.5 M HCl(aq) are needed to neutralize a 288 mL solution of 0.25 M NaOH?
its a dilution problem so..
V1M1=V2M2
0.5(V1)=(288)(.25)
144
To determine the volume of the HCl solution needed to neutralize the NaOH solution, we can use the concept of stoichiometry and the balanced chemical equation between HCl and NaOH. The balanced equation is:
HCl + NaOH -> NaCl + H2O
The stoichiometry of the balanced equation tells us that the ratio between HCl and NaOH is 1:1. This means that 1 mole of HCl reacts with 1 mole of NaOH.
To find the number of moles of NaOH in the given solution, we can use the formula:
moles = concentration (M) x volume (L)
moles of NaOH = 0.25 M x 0.288 L = 0.072 moles
Since the ratio between HCl and NaOH is 1:1, the number of moles of HCl required to neutralize the NaOH is also 0.072 moles.
Now, we can calculate the volume of 0.5 M HCl required using the formula:
volume (L) = moles / concentration (M)
volume of HCl = 0.072 moles / 0.5 M = 0.144 L
To convert this volume to milliliters (mL), multiply by 1000:
volume of HCl = 0.144 L x 1000 mL/L = 144 mL
Therefore, 144 mL of 0.5 M HCl solution are needed to neutralize the 288 mL solution of 0.25 M NaOH.