How many mL of 0.750 M HCl are needed to neutralize 14.89 mL of 0.350 M NaOH? ____ mL

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use the equation M1V1 = M2V2

0.8x100= 0.4 x

7.445ml

To determine how many mL of 0.750 M HCl are needed to neutralize 14.89 mL of 0.350 M NaOH, you can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between HCl and NaOH.

The balanced chemical equation for the neutralization reaction is:
HCl + NaOH -> NaCl + H2O

From the balanced equation, we can see that the ratio of HCl to NaOH is 1:1. This means that 1 mole of HCl reacts with 1 mole of NaOH.

First, let's calculate the moles of NaOH using its concentration and volume:
moles of NaOH = concentration of NaOH × volume of NaOH
moles of NaOH = 0.350 M × 14.89 mL (Remember to convert mL to L by dividing by 1000)
moles of NaOH = 0.350 × 0.01489 mol

Since the reaction is 1:1, the moles of HCl required will be the same as the moles of NaOH.

Now, let's calculate the volume of 0.750 M HCl needed using its concentration and the moles of HCl:
moles of HCl = moles of NaOH
moles of HCl = 0.350 × 0.01489 mol

volume of HCl = moles of HCl / concentration of HCl
volume of HCl = (0.350 × 0.01489 mol) / 0.750 M

volume of HCl = 0.005424 L (Remember to convert back to mL by multiplying by 1000)
volume of HCl = 5.424 mL

Therefore, 5.424 mL of 0.750 M HCl are needed to neutralize 14.89 mL of 0.350 M NaOH.