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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​

Respuesta :

jufsanabriasa

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]

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

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) =

0.804g of NaHCO₃ you must add

okpalawalter8

Answer:

  • [tex]0.8237g[/tex] of [tex]NaHCO_3[/tex]  must be added

Explanation:

Second dissociation of [tex]H_2CO_3[/tex] is

[tex]HCO_3^-(aq) <-------> CO_3^2-(aq) + H^+[/tex]

[tex]Ka = [CO_3^{2-}] [H+] / [HCO_3^-] \\\\= 4.7*10^{-11}\\\\pKa = -log(4.7 * 10^{-11}) \\\\= 10.33[/tex]

[tex]pH = pKa + \frac{log([A-]}{[HA])}\\\\required pH = 10.80\\\\10.80 = 10.33 + \frac{log([A-]}{[HA])}\\\\\frac{log([A-]}{[HA])} \\\\= 0.47\\\\\frac{[A-]}{[HA]}\\\\ = 2.951[/tex]

[tex][CO_3^{2-}] / [HCO_3^-] = 2.951[/tex]

[tex][HCO_3^-] = [CO_3^{2-}]/2.951[/tex]

moles of [tex]K_2CO_3[/tex] = [tex]\frac{4}{138.206} = 0.028942mol[/tex]

moles of [tex]CO_3^{2-}[/tex] = [tex]0.028942mol[/tex]

[tex][CO_3^{2-}][/tex] = [tex]\frac{0.028942}{500} * 1000 = 0.05788M[/tex]

[tex][HCO_3^-][/tex] required = [tex]\frac{[CO32-]}{2.951} = \frac{0.05788}{2.951} = 0.01961M[/tex]

moles of [tex]HCO_3^-[/tex] ​​​​​required = [tex]\frac{0.01961}{1000} * 500 = 0.009805mol[/tex]

moles [tex]NaHCO_3[/tex] required = [tex]0.009805mol[/tex]

Therefore,

mass of [tex]NaHCO_3[/tex] required [tex]= 0.009805 * 84.007 = 0.8237g[/tex]

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