How many mL of 0.49 M HCl will neutralize 42 mL of 1.48 M NaOH?
HCl + NaOH ==> NaCl + H2O
mols NaOH = M x L = ?
mols HCl = mols NaOH (from the coefficients in the balanced equation.)
Then M HCl = mols HCl/L HCl. You know mols and M HCl, solve for L and convert to mL.
so then you would take 1.48 M times 42mL, then take that number and divide by .49 M, then convert to mL?
No, it's 1.48 M times 0.042 L and divide by 049, solve and convert to mL.
However, you can save some time if you do it this way.
1.42 x 42 mL and divide by 0.49. That gives the answer in mL and there is no conversion to do.
To find out how many milliliters of 0.49 M HCl are needed to neutralize 42 mL of 1.48 M NaOH, we can use the concept of molar ratios.
The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
From the equation, we can see that the stoichiometric ratio between HCl and NaOH is 1:1. This means that for every 1 molecule of HCl, we will need 1 molecule of NaOH to completely react.
To determine the amount of HCl needed, we can use the following formula:
(amount of HCl) = (amount of NaOH) x (molar concentration of NaOH) / (molar concentration of HCl)
Substituting the given values, we have:
(amount of HCl) = 42 mL x 1.48 M / 0.49 M
Simplifying the expression:
(amount of HCl) = 126.48 mL / 0.49
(amount of HCl) ≈ 258.32 mL
Therefore, approximately 258.32 mL of 0.49 M HCl is needed to neutralize 42 mL of 1.48 M NaOH.