A 26.8 mL sample of an unknown HClO4 solution requires 46.8 mL of 0.101 M NaOH for complete neutralization. What was the concentration of the unknown HClO4 solution? The neutralization reaction is shown below.
HClO4(aq) + NaOH(aq) H2O(l) + NaClO4(aq)
i got .00473 M HClO4 but it was wrong
moles NaOH = M x L = ??
Look at the equation and use the coefficients to convert moles NaOH to moles HClO4. It's 1:1
Then M HClO4 = moles HClO4/L HClO4.
To solve this problem, we can use the concept of stoichiometry and the equation of the neutralization reaction. The goal is to find the concentration of the unknown HClO4 solution.
First, let's set up the balanced equation for the neutralization reaction:
HClO4(aq) + NaOH(aq) → H2O(l) + NaClO4(aq)
From the balanced equation, we can see that the stoichiometric ratio between HClO4 and NaOH is 1:1. This means that for every 1 mole of HClO4, we will need 1 mole of NaOH for neutralization.
Given that the volume of NaOH used is 46.8 mL and its concentration is 0.101 M, we can calculate the number of moles of NaOH:
moles of NaOH = volume of NaOH (L) x concentration of NaOH (M)
= 0.0468 L x 0.101 M
Next, since the stoichiometric ratio between HClO4 and NaOH is 1:1, the number of moles of HClO4 is also equal to the number of moles of NaOH.
moles of HClO4 = moles of NaOH
Now, we need to convert the volume of the HClO4 solution to liters. The volume is given as 26.8 mL, so:
volume of HClO4 solution (L) = 26.8 mL / 1000 mL/L = 0.0268 L
Finally, we can calculate the concentration of the HClO4 solution:
concentration of HClO4 (M) = moles of HClO4 / volume of HClO4 solution (L)
= moles of NaOH / volume of HClO4 solution (L)
= (0.0468 L x 0.101 M) / 0.0268 L
Now, let's do the calculation:
concentration of HClO4 (M) = (0.0468 L x 0.101 M) / 0.0268 L
≈ 0.176 M
So, the concentration of the unknown HClO4 solution is approximately 0.176 M, not 0.00473 M as you initially calculated.
Remember to carefully double-check your calculations to ensure accuracy.