(A) In a solution 15.0 mL of 4.00 M HCl is titrated with 0.750 M NaOH. How many
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Question
(A) In a solution 15.0 mL of 4.00 M HCl is titrated with 0.750 M NaOH. How many milliliters of NaOH are required to reach the equivalent point and a pH of 7.00?
(b) In a solution 15.0 mL of 4.00 M HCl is titrated with 0.750 M NaOH. What is the pH when the volume of NaOH added is 0.20 mL less than the volume required to reach the equivalence point is.
(c) In a solution 15.0 mL of 4.00 M HCl is titrated with 0.750 M NaOH. The pH when the volume of NaOH added is 0.20 mL more than the volume required to reach the equivalence point is:
Explanation / Answer
Answer.
a.) We have,
For HCl, M1 = 4 M, V1 = 15 mL
For NaOH, M2 = 0.75 M, V2 = ?
Using the equation of moles equalization, we have
M1V1 = M2V2
15*4 = 0.75 * V2
V2 = 80 mL.
Hence, 80 milliliters of NaOH are required to reach the equivalent point and a pH of 7.00.
b.) Now,
Volume of NaOH (V2) = 80 - 0.2 = 79.8 mL
Therefore,
Moles of HCl = M1*V1 = 15*4 = 60 millimoles
Moles of NaOH = M2*V2 = 0.75*79.8 = 59.85 millimoles
Total Volume = V1 + V2 = 15 + 79.8 = 94.8 mL = 0.0948 L
Extra moles of HCl = moles of [H+] = 60-59.85 = 0.15 millimoles
Concentration of [H+] = 0.15 * 10-3 / 0.0948 = 0.00158227848 M
Hence,
pH = - log([H+]) = -log(0.00158227848) = 2.80071708.
c.)
Now,
Volume of NaOH (V2) = 80 + 0.2 = 80.2 mL
Therefore,
Moles of HCl = M1*V1 = 15*4 = 60 millimoles
Moles of NaOH = M2*V2 = 0.75*80.2 = 60.15 millimoles
Total Volume = V1 + V2 = 15 + 80.2 = 95.2 mL = 0.0952 L
Extra moles of NaOH = moles of [OH-] = 60.15-60 = 0.15 millimoles
Concentration of [H+] = 0.15 * 10-3 / 0.0952 = 0.00157563025 M
Hence,
pH = 14 - pOH = 14 + log([OH-]) = 14 + log(0.00157563025) = 11.1974543.
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