a 50.0ml sample of a 1.00m naoh is titrated with 1.00m hcl what is the ph What i
ID: 1007415 • Letter: A
Question
a 50.0ml sample of a 1.00m naoh is titrated with 1.00m hcl what is the phWhat is the ph after the addition of 10.00ml of HCl?
What is the ph at the equivalence point?
What volume of HCl is required to reach equivalence point?
a 50.0ml sample of a 1.00m naoh is titrated with 1.00m hcl what is the ph
What is the ph after the addition of 10.00ml of HCl?
What is the ph at the equivalence point?
What volume of HCl is required to reach equivalence point?
a 50.0ml sample of a 1.00m naoh is titrated with 1.00m hcl what is the ph
What is the ph after the addition of 10.00ml of HCl?
What is the ph at the equivalence point?
What volume of HCl is required to reach equivalence point?
Explanation / Answer
mols NaOH initially = M x L = 50.0 ml sample of a 1.00 M NaOH = 0.05 L X 1M = 0.05 Moles
mols HCl added = M x L = 10.00 ml of HCl = 0.01 X 1 = 0.01 Moles
mols NaOH remaining = mols NaOH - mols HCl. = 0.05-0.01 = 0.04 Moles in 60 mL
Therefore [OH] = 0.67 Molar solution
pOH = -log (OH^-) = - log 0.67 = 0.174
Then pH + pOH = pKw = 14.
pH = 13.8
2. In the case of titration of strong acid (HCl) with strong base (NaOH) (or strong base with strong acid) there is no hydrolysis and solution pH is neutral - 7.00 (at 25°C).
3. since both NaOH and HCl are 1 M solution the volume of HCl at equivalence point is 50 mL itself.
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