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When 0.4 g of NaOH is dissolved in one litre of solution, the pH of the solution is: $$NaOH\rightarrow { Na }^{ + }+{ OH }^{ - }$$ Strong base completely dissociates, therefore $$\left[ { OH }^{ - } \right] =\frac { 0.4 }{ 40 } mol/L={ 10 }^{ -2 }M$$ Since, $$\left[ { OH }^{ - } \right] \left[ { H }^{ + } \right] ={ 10 }^{ -14 }M\\ \left[ { H }^{ + } \right] ={ 10 }^{ -12 }M\\ pH=-log\left( \left[ { H }^{ + } \right] \right) =12$$ Therefore option A is correct.
Now the #"moles of solute"# are a constant. The volume of solution MAY change substantially with increasing or decreasing temperature. In some calculations #"molality"# is used in preference, which is defined by the quotient.... #"molality"="moles of solute"/"kilograms of solvent"# ....this is temperature independent, and at lower concentrations, #"molarity"-="molality"#.
To get the molarity, you divide the moles of solute by the litres of solution. #"Molarity" = "moles of solute"/"litres of solution"# For example, a 0.25 mol/L NaOH solution contains 0.25 mol of sodium hydroxide in every litre of solution. To calculate the molarity of a solution, you need to know the number of moles of solute and the total volume of the solution. To calculate molarity:
EXAMPLE: What is the molarity of a solution prepared by dissolving 15.0 g of NaOH in enough water to make a total of 225 mL of solution? Solution: 1 mol of NaOH has a mass of 40.00 g, so #"Moles of NaOH" = 15.0 cancel("g NaOH") × "1 mol NaOH"/(40.00 cancel("g NaOH")) = "0.375 mol NaOH"# #"Litres of solution" = 225 cancel("mL soln") × "1 L soln"/(1000 cancel("mL soln")) = "0.225 L soln"# #"Molarity" = "moles of solute"/"litres of solution" = "0.375 mol"/"0.225 L" = "1.67 mol/L"# Some students prefer to use a "molarity triangle". It summarizes the molarity formulas as #"Moles" = "molarity × litres"# #"Molarity" = "moles"/"litres"# #"Litres" = "moles"/"molarity"#
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