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.

1) pH=3

2) its colour is blue in basic solution

3) KOH(aq) +HCl(aq) > KCl(aq) + H2O(l)

4) 0.02 M

5) increase in solution concentration (number of ions present in solution) increases its electrical conductivity

Explanation:

1) A solution ten times less acidic than the HCl(aq) will have a concentration of 0.001M

Hence from pH= -log[H^+].

pH= - log(0.001)

pH= 3

2) its colour is blue in basic solution

3) KOH(aq) +HCl(aq) > KCl(aq) + H2O(l)

4)

from

CA= concentration of acid= 0.010M

CB= concentration of base=

VA= volume of acid= 15.0mL

VB= volume of base= 7.5mL

NA= number of moles of acid= 1

NB= number of moles of base= 1

CAVA/CBVB=NA/NB

CAVANB=CBVBNA

CB= CAVANB/VB NA

CB= 0.01 ×15.0 × 1/ 7.5 ×1

CB= 0.02 M

5) The more concentrated the solution is, the more the number ions present and the better the conductivity of the solution. Hence the more concentrated 1.0M HCl(aq) contains more ions and is a better conductor of electricity than 0.01M HCl(aq)