How much 0.150 M sodium hydroxide would be required to just neutralize 15 mL of 0.175M hydrochloric acid?
15mL x 0.150M/0.175M=12mL
correct!
No.
mL NaOH x M NaOH = mL HCl x M HCl
I think you have some numbers reversed.
To calculate the amount of sodium hydroxide required to neutralize 15 mL of hydrochloric acid, you need to use the equation for stoichiometry.
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is:
NaOH + HCl -> NaCl + H2O
From the balanced equation, you can see that the molar ratio between NaOH and HCl is 1:1. This means that for every one mole of NaOH, you need one mole of HCl.
Given:
Volume of HCl solution (VHCl) = 15 mL
Concentration of HCl solution (CHCl) = 0.175 M
Concentration of NaOH solution (CNaOH) = 0.150 M
To find the amount of sodium hydroxide needed, you can use the following formula:
VNaOH = VHCl * CHCl / CNaOH
Substituting the given values:
VNaOH = 15 mL * 0.175 M / 0.150 M
Simplifying:
VNaOH = 17.5 mL / 0.150 M
VNaOH ≈ 116.67 mL
Therefore, approximately 116.67 mL of 0.150 M sodium hydroxide would be required to just neutralize 15 mL of 0.175 M hydrochloric acid.