An unknown amount of water is mixed with
350 mL of a 6 M solution of NaOH solution. A 75 mL sample of the resulting solution is titrated to neutrality with 52.5 mL of 6 M HCl. Calculate the concentration of the diluted NaOH solution.
Answer in units of M
What are the steps to doing this... your answer is correct, I just want to know because I could really help me out with my chemistry test next week.
0.052L*6moles/1liter gives you 0.312 moles of HCL, meaning that is also the moles of NaOH. It gives you 0.075 L or 75 mL, 0.312/0.075 is 4.2M
I think.
HCl + NaOH ==> NaCl + H2O
mols HCl = M x L = ?
mols NaOH = mols HCl
M NaOH = mols NaOH/L NaOH
I believe that works out to be about 4.2M but check my work.
To find the concentration of the diluted NaOH solution, we can use the concept of moles and the principle of equivalence in acid-base reactions.
First, let's calculate the number of moles of HCl used in the titration. We know that the concentration of the HCl solution is 6 M, and the volume used is 52.5 mL.
Number of moles of HCl = Concentration × Volume
Number of moles of HCl = 6 M × 52.5 mL
Next, let's find the number of moles of NaOH that reacted with the HCl. According to the balanced chemical equation, 1 mole of HCl reacts with 1 mole of NaOH in a 1:1 ratio.
Since the solution is neutral when the reaction is complete, the number of moles of HCl used will be equal to the number of moles of NaOH present in the 75 mL sample.
Number of moles of NaOH = Number of moles of HCl
Now, let's find the concentration of the diluted NaOH solution in the original mixture. We know that the 75 mL sample is a portion of the overall mixture, and we need to consider the dilution by the unknown amount of water.
Since the moles of NaOH are the same in the diluted solution and the original mixture, we can set up the following equation:
Moles of NaOH in the diluted solution = Moles of NaOH in the original mixture
Concentration of NaOH in the diluted solution × Volume of the diluted solution = Concentration of NaOH in the original mixture × Volume of the original mixture
We have the following values:
Concentration of NaOH in the diluted solution = unknown (let's represent it as x M)
Volume of the diluted solution = 75 mL
Concentration of NaOH in the original mixture = 6 M
Volume of the original mixture = 350 mL
Plugging in these values into the equation:
x M × 75 mL = 6 M × 350 mL
Simplifying the equation:
x M = (6 M × 350 mL) / 75 mL
Calculating the concentration of the diluted NaOH solution:
x M ≈ 28 M
Therefore, the concentration of the diluted NaOH solution is approximately 28 M.