A student dissolves 5.00 g calcium hydroxide to make 100.0 mL of solution. If 63.5 mL of this solution is required to neutralize 150.0 mL of a hydrofluoric acid solution, what is the concentration of the original HF solution? (The molar mass of Ca(OH)2 is 74.10 g/mol.)
I need help with this asap please.
5.00 g is dissolved in 100.0 mL
So,
63.5 mL will contain, 5*63.5/100 = 3.175 g
This should be neutralised
we have the Balanced chemical equation as:
Ca(OH)2 + 2 HF ---> CaF2 + 2 H2O
Molar mass of Ca(OH)2 = 1*MM(Ca) + 2*MM(O) + 2*MM(H)
= 1*40.08 + 2*16.0 + 2*1.008
= 74.096 g/mol
mass of Ca(OH)2 = 3.175 g
we have below equation to be used:
number of mol of Ca(OH)2,
n = mass of Ca(OH)2/molar mass of Ca(OH)2
=(3.175 g)/(74.096 g/mol)
= 4.285*10^-2 mol
From balanced chemical reaction, we see that
when 1 mol of Ca(OH)2 reacts, 2 mol of HF reacts
mol of HF reacted = (2/1)* moles of Ca(OH)2
= (2/1)*4.285*10^-2
= 8.57*10^-2 mol
This is number of moles of HF
volume , V = 150.0 mL
= 0.15 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 8.57*10^-2/0.15
= 0.5713 M
Answer: 0.571 M
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