Question

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Answer 1

molarity of HCl =0.150M

1000ml of solution contain 0.150 moles of HCl

1 ml of solution contain 0.150x10^-3moles of HCl

15ml of solution contains 2.25x10^-3 moles of HCl

a) molarity of NaOH = 0.150M

  1000ml of solution contain 0.150 moles of NaOH

1 ml of solution contain 0.150x10^-3moles of NaOH

15ml of solution contains 2.25x10^-3 moles of NaOH

now we have the reaction

NaOH +HCl   $\rightarrow$ H2O + NaCl

so from the reaction, we say that

1 mole of NaOH required to completely neutralize the 1 mole of HCl

A 2.25x10^-3 mole of NaOH will completely neutralize the 2.25x10^-3 mole of HCl

so the solution will be neutral and the pH for a neutral solution is 7.

b)

molarity of NaOH = 0.200M

  1000ml of solution contain 0.200 moles of NaOH

1 ml of solution contain 0.200x10^-3moles of NaOH

25ml of solution contains 5x10^-3 moles of NaOH

now we have the reaction

NaOH +HCl   $\rightarrow$ H2O + NaCl

1 mole of NaOH required to completely neutralize the 1 mole of HCl

The 2.25x10^-3 mole of HCl is completely neutralized by  2.25x10^-3 mole of NaOH

so here NaOH is an excess reagent

remaining mole of NaOH = 5x10^-3 - 2.25x10^-3 = 2.75x10^-3 moles

so the concentration of NaOH = moles of NaOH / total volume of solution

= 2.75x10^-3 / (25+15)x10^-3

= 0.07M

pOH = -log([OH^-]) = -log(0.07) =1.15

pH = 14 - 1.15 = 12.85