1. write the reaction of HI in water. 2 .0251 mol of HI is dissolved in enough w

Question

1. write the reaction of HI in water. 2 .0251 mol of HI is dissolved in enough water to make 1.50 L of solution. Calculate the concentrations (molarities) of HI, H3O+ and I- in the solution after the reaction is complete. 3. calculate the pH of the solution in part 2. 3 Write the reaction of chloric acid in water. 3a in your reaction for part a., circle the conjugate base for chloric acid (HCLO3). 3c write the reaction of the conjugate base from part b. with water. d write the reaction of chloric acid with pyridine (C5H5N--a weak base). e in your reaction for part d., circle the conjugate acid of pyridine.

Explanation / Answer

1.

HI(aq)+H2O(l) ---> H3O+(aq) +I-(aq)

2)

.0251 mol of HI is dissolved in enough water to make 1.50 L of solution

0.0251 mole in 1.5 L since it is strong acid all HI will excist as [H3O]+ and I-

SO [H3O]+ concentration = 0.0251/1.5*1

= 0.016733molar

I- Concentration = 0.016733molar

HI concentration = 0

3)

pH of solution = - log [H+] = -log (0.016733) = 1.77

4)

chloric acid in water

HClO3 is an acid, meaning it donates hydrogen ions to willing recipients (bases). H2O can act as an acid or, in cases like this, as a base. As a base water receives hydrogen ions, forming H3O^+1 ions.

HClO3(aq) + H2O(l) --> ClO3^-1(aq) + H3O^+1(aq)

conjugate base for chloric acid (HCLO3) =  ClO3^-1

reaction of the conjugate base

ClO3^-1(aq) + H2O ----> HClO3 + OH-

5) reaction of chloric acid with pyridine

HClO3(aq) + C5H5N -----> C5H5N(+)H + ClO3-

Conjugate acid of pyridine =  C5H5N(+)H

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