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

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) =

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3