text: Question
  
  # The temperature for each solution is carried out at approximately 297 K where Kw=1.00×10−14 Part A...
  
  The temperature for each solution is carried out at approximately 297 K where Kw=1.00×10−14
  
  Part A
  
  0.25 g of hydrogen chloride (HCl) is dissolved in water to make 8.0 L  of solution. What is the pH of the resulting hydrochloric acid solution?
  
  Part B
  
  0.90 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 6.5 L of solution. What is the pH of this solution?
  
  Express the pH numerically to two decimal places
  
  part C
  
  What is the pOH of the solution in part B ?.
  
  .
  
  Part A :
  
  Number of moles of HCl , n = mass/molar mass
  
  = 0.25 g / ( 36.5(g/mol)
  
  = 6.85x10-3 mol
  
  So concentration of HCl = number of moles / volume in L
  
  = 6.85x10-3 mol / 8.0 L
  
  = 8.56x10-4 M
  
  HCl ---> H+ + Cl-
  
  1 mole of HCl contains 1 mole of H+
  
  So [H+] = [HCl] = 8.56x10-4 M
  
  pH = - log[H+]
  
  = - log(8.56x10-4 )
  
  = 3.07
  
  Part B :
  
  Molarity of NaOH = ( mass/molar mass) / volume in L
  
  = ( 0.90 g / 40(g/mol) ) / 6.5 L
  
  = 3.46x10-3 M
  
  1 mole of NaOH contains 1 mole of OH-
  
  [OH-] = [NaOH] = 3.46x10-3 M
  
  pOH = - log (3.46x10-3 )
  
  = 2.46
  
  pH = 14-pOH
  
  = 14-2.46
  
  = 11.54
  
  Part C :
  
  pOH = 2.46
  
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