Question

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.

Answer #1

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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