2. Calculate the approximate weight of KHP required so that about 35 mL of 0.1 N

Question

2. Calculate the approximate weight of KHP required so that about 35 mL of 0.1 N sodium hydroxide will be used in a titration. (Ew. KHP-204.23 g/equiv.)-follow regular sig fig rules! Amount Record this amount in your lab notebook! 3. A 0.6237 g sample of KHP with a purity of 99.99% is titrated with 42.34 mL of NaOH solution. From the data given, calculate the normality of base. Round your answer to the appropriate number of significant figures based upon a precision of 1 ppth. ppth sig figs!! NaOHconc Revised Sp 2016 NF/AEM/RLG Page 11 of 12 Standardization of NaOH

Explanation / Answer

(2) Equivalent mass of KHP (HKC8H4O4) = 204.23 g/mol

HKC8H4O4 + NaOH -----> NaKC8H4O4 + H2O

Moles of KHP required to neutralize 0.1 N (g/mol) NaOH = NNaOHVNaOH = (0.1/1000)(g/ml)x35 ml

Moles of KHP = 0.0035 mol

Mass of KHP = 0.0035 mol x204.23 g/mol = 0.715 g

(3) Using the similar approach as mentioned above:

Moles of KHP required to neutralize n N (g/mol) NaOH = NNaOHVNaOH = (n/1000)(g/ml)x42.34 ml

Moles of KHP = 0.6237 / 204.23 =0.00305 mol

n = 0.00305 x 1000 / 42.34 = 0.072 Normal

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