8 gram NaOH is dissolved in water and the aqueous solution is made to 5 litres.Find pH of this solution. Solve the given problem.
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Given, mass of NaOH, w = 8gm
volume of solution , V = 5 Litres
molar mass of NaOH, M = 40 gm/mol
so, molarity of NaOH = no of moles of NaOH/volume of solution in Litres
= given weight of NaOH /molar mass of NaOH × volume of solution in L
= w/(M × V) = 8/(40 × 5)
= 8/200 = 40/1000
= 0.04 M
hence, concentration of OH- = 0.04M
so, [OH-] = 0.04 M
we know, according to Arrhenius concepts
POH = -log10[OH-]
= -log10(0.04)
= -log10(4 × 10^-2)
= 2 - log₁₀4
= 2 - 0.6020 = 1.39
hence, pOH = 1.39
we know, PH + POH = 14
so, PH = 14 - 1.39 = 12.61
therefore , PH of given solution = 12.61
volume of solution , V = 5 Litres
molar mass of NaOH, M = 40 gm/mol
so, molarity of NaOH = no of moles of NaOH/volume of solution in Litres
= given weight of NaOH /molar mass of NaOH × volume of solution in L
= w/(M × V) = 8/(40 × 5)
= 8/200 = 40/1000
= 0.04 M
hence, concentration of OH- = 0.04M
so, [OH-] = 0.04 M
we know, according to Arrhenius concepts
POH = -log10[OH-]
= -log10(0.04)
= -log10(4 × 10^-2)
= 2 - log₁₀4
= 2 - 0.6020 = 1.39
hence, pOH = 1.39
we know, PH + POH = 14
so, PH = 14 - 1.39 = 12.61
therefore , PH of given solution = 12.61
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