June 2019 Chemistry Regents #81-85

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Question 81-85

 

 

Base your answers to questions 81-85

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 KOH(aq) is exactly neutralized by 15.0milliliters of the 0.010 M HCl(aq). During this laboratory activity, appropriate safety equipment is used and safety procedures are followed.

 

81.  Determine the pH value of a solution that is ten times less acidic than HCl (aq) solution.

Answer-->    3

82.  State the color of the indicator bromocresol green if it is added to a sample of the KOH(aq).

Answer-->   blue

83.  Complete the equation for the reaction

HCl + KOH-->  ______   +   ______

Answer-->  HCl + KOH-->  KCl  + H2O

H2O....HOH..OH2..(ugg)...KCl...ClK..(ugg)

84.  Show the correct numerical setup to calculate the molarity of the KOH solution.

Answer-->   MV=MV

0.010M(15.0ml)=M(7.5mL)  .....no units needed...

do not solve

85.  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.

Answer-->  There are more aqueous ions

alides

see Table F halides

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