How many mL of .175M H3PO4 are needed to neutralize 15.00mL of .33M NaOH?
Write the equation and balance it.
mols NaOH = M x L = ?
Using the coefficients, convert mols NaOH to mols H3PO4.
The M H3PO4 = mols/L. You know mols and M, solve for L and convert to mL.
Looks like the M1V1 = M2V2 to solve the problem....?
No. M1V1 = M2V2 won't do it. That works ONLY for solutions that are 1:1 in the reaction. This one is 3NaOH to 1 H3PO4.
To determine the number of milliliters (mL) of 0.175M H3PO4 that are needed to neutralize 15.00mL of 0.33M NaOH, we can use the concept of stoichiometry.
First, let's write the balanced chemical equation for the neutralization reaction between H3PO4 and NaOH:
H3PO4 + 3NaOH -> Na3PO4 + 3H2O
From the equation, we can see that 1 mole of H3PO4 reacts with 3 moles of NaOH. Therefore, the stoichiometric ratio is 1:3.
Now, let's calculate the number of moles of NaOH used by using the given volume and concentration:
moles of NaOH = volume (in L) x concentration (in M)
= 0.015L x 0.33M
= 0.00495 moles NaOH
According to the stoichiometric ratio, this means that we will need three times the number of moles of H3PO4 to neutralize the NaOH:
moles of H3PO4 = 3 x moles of NaOH
= 3 x 0.00495 moles NaOH
= 0.01485 moles H3PO4
Finally, we can determine the required volume (in mL) of 0.175M H3PO4 by using the moles of H3PO4 and its concentration:
volume (in L) = moles / concentration
= 0.01485 moles H3PO4 / 0.175M
= 0.08457 L
volume (in mL) = 0.08457 L x 1000 mL/L
= 84.57 mL
Therefore, approximately 84.57 mL of 0.175M H3PO4 are needed to neutralize 15.00 mL of 0.33M NaOH.