Calculate the amount (volume) of 0.135 M HClO4 solution required to neutralize 57.00 mL of .089 M NaOH
This is a 1:1 acid/base reaction.
HClO4 + NaOH ==> NaClO4 + H2O
mL1 x M1 = mL2 x M2
Substitute and solve for the unknown.
To calculate the volume of HClO4 solution required to neutralize a certain volume of NaOH solution, we first need to determine the balanced chemical equation for the reaction between HClO4 and NaOH:
HClO4 + NaOH → NaClO4 + H2O
In this reaction, 1 mole of HClO4 reacts with 1 mole of NaOH to produce 1 mole of NaClO4 and 1 mole of water.
Next, we need to determine the number of moles of NaOH that is present in 57.00 mL of 0.089 M NaOH solution. To do this, we use the equation:
moles = volume (L) × concentration (mol/L)
First, convert the volume from mL to L:
57.00 mL ÷ 1000 = 0.057 L
Then, calculate the number of moles:
moles of NaOH = 0.057 L × 0.089 mol/L = 0.005073 mol
Now that we know the number of moles of NaOH, we can determine the number of moles of HClO4 required to neutralize it. From the balanced chemical equation, we know that the stoichiometry is 1:1 between NaOH and HClO4.
Therefore, the number of moles of HClO4 required is also 0.005073 mol.
Finally, we need to calculate the volume of 0.135 M HClO4 solution required to provide 0.005073 mol of HClO4. We can use the same formula as above:
volume (L) = moles ÷ concentration (mol/L)
volume of HClO4 = 0.005073 mol ÷ 0.135 mol/L = 0.0376 L
Convert the volume back from liters to milliliters:
0.0376 L × 1000 = 37.6 mL
Therefore, the amount (volume) of 0.135 M HClO4 solution required to neutralize 57.00 mL of 0.089 M NaOH is approximately 37.6 mL.