Suppose you titrate 80.0 mL of 2.00 M NaOH with 20.0 mL of 4.00 M HCl. What is the final concentration of OH− ions.
2 answers:
Total volume = 80 mL + 20 mL = 100 mL = 0.10 L.
NaOH + HCl ----> NaCl + H2O.
0.08 L * 2.00 mol NaOH/L = 0.16 moles NaOH.
0.02 L * 4.00 mol HCl/L = 0.08 moles HCl.
0.16 - 0.08 = 0.08. There are 0.08 moles NaOH XS.
0.08 moles NaOH produces 0.08 moles OH-.
Concentration OH- = 0.08 moles / 0.10 L = 0.8 M.
Answer: The final concentration of
is 0.8 M.
Explanation: 


0.06 moles of NaOH will give 0.06 moles of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)



0.08 moles of HCl will give 0.08 moles of ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)

0.08 moles of
will react with 0.08 moles of
and (0.16-0.08)= 0.08 moles of
will be left in 100 ml of solution.
Thus Molarity of ![[OH^]-=\frac{moles}{\text {Volume in L}}=\frac{0.08}{0.1}=0.8M](https://tex.z-dn.net/?f=%5BOH%5E%5D-%3D%5Cfrac%7Bmoles%7D%7B%5Ctext%20%7BVolume%20in%20L%7D%7D%3D%5Cfrac%7B0.08%7D%7B0.1%7D%3D0.8M)
You might be interested in
It decreases i guess. just make sure,though i'm mostly correct.
Compound made up of hydrogen and carbon
It could explode
or blow up
hope it helped