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KiRa [710]
3 years ago
5

How many mL of 0.125 M Ba(OH)2 would be required to completely neutralize 75.0 mL of 0.845 M HCl? What is the pH of the solution

at the equivalence point?
Chemistry
1 answer:
torisob [31]3 years ago
8 0

Answer:

253.5mL of Ba(OH)₂ are required to neutralize the HCl solution

The pH at equivalence point is = 7

Explanation:

The reaction of Ba(OH)₂ with HCl is:

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O

First, we need to find moles of HCl. With these moles and the chemical equation we can find the moles of Ba(OH)₂ and the volume required:

<em>Moles HCl:</em>

75.0mL = 0.075L * (0.845mol / 1L) = 0.063375moles HCl

<em>Moles Ba(OH)₂:</em>

0.06338moles HCl * (1mol Ba(OH)₂ / 2mol HCl) = 0.03169 moles Ba(OH)₂

<em>Volume of the 0.125M Ba(OH)₂:</em>

0.03169 moles Ba(OH)₂ * (1L / 0.125mol) = 0.2535L are required =

253.5mL of Ba(OH)₂ are required to neutralize the HCl solution

As the titration was of a strong acid, HCl, with a strong base, Ba(OH)₂, the pH at equivalence point is = 7

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First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

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0.107M=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{82g/mol\times 125mL}

\text{Mass of }NaCH_3CO_2=1.097g

Therefore, the mass of sodium acetate is, 1.097 grams.

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