Isocyanic acid (HNCO) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation 2 NaOCN ( s ) + H 2 C 2 O 4 ( s ) → 2 HNCO ( l ) + Na 2 C 2 O 4 ( s ) Upon isolating pure HNCO( l ), an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a l00.-mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and H 2 C 2 O 4 , assuming all of the HNCO produced is dissolved in solution? ( K a of HNCO = 1.2 × l0 −4 .)

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Interpretation Introduction Interpretation: The pH of a 100 mL solution of HNCO is to be stated. Concept introduction: A limiting reagent is the substance which is completely used up or reacted in a chemical reaction. The limiting reagent determines the amount of product formed in a chemical reaction. Explanation Explanation The concentration of HNCO is 1 . 5 4 M _ . The complete balanced reaction is, 2 NaOCN + H 2 C 2 O 4 → 2 HNCO + Na 2 C 2 O 4 Thus, 2 moles of NaOCN reacts with 1 mole of H 2 C 2 O 4 to form 2 moles of HNCO . Number of moles of NaOCN = Given mass moleculer mass = 10 65 = 0.154 moles Number of moles of H 2 C 2 O 4 = Given mass Moleculer mass = 10 90.034 = 0.111 moles Thus, H 2 C 2 O 4 requires 0.222 moles of NaOCN but the present amount of NaOCN is 0.154 moles . Hence, NaOCN is the limiting reagent. 2 moles of NaOCN form 2 moles of HNCO . 0.154 moles of NaOCN form 0.154 moles of HNCO . The concentration is calculated by the formula, Concentration of HNCO = Number of moles Volume of solution Substitute the value of the number of moles and the volume of the solution in the above expression. Concentration of HNCO = Number of moles Volume of solution = 0.154 0.1 L = 1 . 5 4 M _ The value of [ H + ] is 0 . 0 1 3 5 M _ . The ICE (Initial Change Equilibrium) table of the dissociation of HCNO is, HCNO + H 2 O ⇌ H 3 O + + NCO − Initial ( M ) : 1

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Interpretation Introduction

Interpretation: The pH of a 100 mL solution of HNCO is to be stated.

Concept introduction: A limiting reagent is the substance which is completely used up or reacted in a chemical reaction. The limiting reagent determines the amount of product formed in a chemical reaction.

Explanation of Solution

Explanation

The concentration of HNCO is 1.54 M_.

The complete balanced reaction is,

2NaOCN+H2C2O4→2HNCO+Na2C2O4

Thus, 2 moles of NaOCN reacts with 1 mole of H2C2O4 to form 2 moles of HNCO.

Number of moles of NaOCN=Given massmoleculer mass=1065=0.154 moles

Number of moles of H2C2O4=Given massMoleculer mass=1090.034=0.111 moles

Thus, H2C2O4 requires 0.222 moles of NaOCN but the present amount of NaOCN is 0.154 moles. Hence, NaOCN is the limiting reagent.

2 moles of NaOCN form 2 moles of HNCO.

0.154 moles of NaOCN form 0.154 moles of HNCO.

The concentration is calculated by the formula,

Concentration of HNCO=Number of moles Volume of solution

Substitute the value of the number of moles and the volume of the solution in the above expression.

Concentration of HNCO=Number of moles Volume of solution=0.1540.1 L=1.54 M_

The value of [H+] is 0.0135 M_.

The ICE (Initial Change Equilibrium) table of the dissociation of HCNO is,

HCNO+H2O⇌H3O++NCO−Initial(M) :1

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