Suppose you titrate 80.0 mL of 2.00 M NaOH with 20.0 mL of 4.00 M HCl....

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Suppose you titrate 80.0 mL of 2.00 M NaOH with 20.0 mL of 4.00M HCl. What is the final concentration of OH- ions.

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Number of moles of NaOH = M(NaOH) * V(NaOH)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 M * 0.08 L
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.16 mol
Number of moles of HCl = M(HCl) * V(HCl)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4 M * 0.02 L
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.08 mol
HCL + NaOH -->NaCL + H2O
0.08 mol of each will react and neutralise each other

After Reaction :
Number of molesof NaOH remaining = 0.16 - 0.08 = 0.08 mol
Total volume= 80 mL+ 20mL =100 mL = 0.1 L
[OH-]=[NaOH]=Number of moles /Volume
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.08 / 0.1
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.8 M
Answer: final concentration of OH- ions= 0.8 M

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