Chap 4 question 13 box. How many grams of KHP are needed to neutralize 75.47 mL

Question

Chap 4 question 13 box. How many grams of KHP are needed to neutralize 75.47 mL of a 0.1041 M NaOH solution? (molar mass of KHP = 204.2 g/mol) Chap 4 question 13 box. How many grams of KHP are needed to neutralize 75.47 mL of a 0.1041 M NaOH solution? (molar mass of KHP = 204.2 g/mol) box. How many grams of KHP are needed to neutralize 75.47 mL of a 0.1041 M NaOH solution? (molar mass of KHP = 204.2 g/mol) box. How many grams of KHP are needed to neutralize 75.47 mL of a 0.1041 M NaOH solution? (molar mass of KHP = 204.2 g/mol)

Explanation / Answer

The reaction taking place during the titration is:

KHP(aq) + NaOH(aq) --> H2O(l) + KNap(aq)

molar mass(KHP) = 204.2 g/mol

for standardisation,

mol of KHP = mol of NaOH

or,

mass(KHP)/molar mass(KHP) = M(NaOH)*V(NaOH)

mass/ (204.2 g/mol) = 0.1041 M * 0.07547 L

mass/ (204.2 g/mol) = 7.856*10^-3 mol

mass = 1.60 g

Answer: 1.60 g

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