1.c) 1 to 1.2 gram samples of KHP (204.22g/mol, potassium hydrogen phthalate, no
ID: 712169 • Letter: 1
Question
1.c)1 to 1.2 gram samples of KHP (204.22g/mol, potassium hydrogen phthalate, not potassium hydrogen phosphorous) will be used to standardize the NaOH solution.Approximately what volume of the NAOH solutions will be requires to titrate the KHP?
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Prelaboratory Problems - Experiment 21 - Acids and Bases: Reactions and Standardization The solutions to the starred problems are in Appendix 4 Option 1-a of the instructions suggests the dissolving of about 10 g of NaOH in 0.95 L of water to prepare the solution to be standardized. Calculate the approximate concentration of this solution. 1. a. o.aSL Nadu Option 1-b of the instructions suggests that about 41 mL of 6 M NaOH be diluted to about 0.95 L of water to prepare the solution to be standardized. Calculate the approximate concentration of this solution. b. Mi- M- 0.24M 04.22 mot,potassid m hydrogen phthalate, not potassiunm 1 to 1.2 gram samples of KHP (2 hydrogen phosphorous) will be used to standardize the NaOH solution. Approximately what volume of the NaOH solution will be required to titrate the KHP? c.* 0.0217
Explanation / Answer
Ans. Moles of KHP taken = Mass / Molar mass
= 1.1 g / (204.22 g / mol)
= 0.0053863 mol
# Balanced reaction: KHC8H4O4(aq) + NaOH(aq) -------> KNaC8H4O4(aq) + H2O(l)
According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol KHP (KHC8H4O4).
That is,
Moles of NaOH consumed = Moles of KHP neutralized
Hence, Moles of NaOH consumed = 0.0053863 mol
Now,
Required volume of NaOH solution = Moles of NaOH / Molarity
= 0.0053863 mol / 0.248 M
= 0.0053863 mol / (0.248 mol/ L)
= 0.02172 L
= 21.72 mL
Therefore, required volume of NaOH solution = 21.72 mL
# Note: You calculation is correct.
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