In a laboratory investigation, an HCl(aq) solution with a pH value of 2 is used to determine the molarity of a KOH(aq) solution. A 7.5-milliliter sample of the KOH(aq) is exactly neutralized by 15.0 milliliters of the 0.010 M HCl(aq). During this laboratory activity, appropriate safety equipment is used and safety procedures are followed. 1. Determine the pH value of a solution that is ten times less acidic than HCl (aq) solution.2. State the color of the indicator bromocresol green if it is added to a sample of the KOH(aq).3. Complete the equation for the reaction: HCl + KOH--> + .4. How the correct numerical setup to calculate the molarity of the KOH solution. 5. Explain in terms of aqueous ions, why 15.0mL of 1.0M HCl(aq) is a better conductor of electricity than 15.0mL of 0.010M HCl(aq) solution.

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answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole