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DENIUS [597]
2 years ago
10

how many grams of nahco3 (fm84.007) must be added ro 4.00g of k2co3 (fm 138.206) to give a ph of 10.80 in 500ml of water​

Chemistry
1 answer:
Anna11 [10]2 years ago
6 0

Answer:

0.804g of NaHCO₃ you must add

Explanation:

pKa of HCO₃⁻/CO₃²⁻ is 10.32.

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

<em>Where [HA] is concentration of acid (HCO₃⁻) and [A⁻] is concentration of conjugate acid (CO₃²⁻).</em>

Moles of CO₃²⁻ = K₂CO₃ are:

4.00g ₓ (1mol / 138.206g) = 0.0289 moles CO₃²⁻

Replacing:

10.80 = 10.32 + log [0.0289] / [HCO₃⁻]

[HCO₃⁻] = 0.009570 moles you need to add to obtain the desire pH

As molar mass of NaHCO₃ is 84.007g/mol, mass of NaHCO₃ is:

0.009570 moles ₓ (84.007g / mol) =

<h3>0.804g of NaHCO₃ you must add</h3>

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      <em><u>calculation</u></em>

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        that  is  for  Li= 0.624/0.623=  1

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<h3>Therefore the formula=LiClO4.3H2O</h3><h3 />
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