How many grams of dry NH4Cl need to be added to 2.00 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.63? Kb for ammonia is 1.8×10−5.

Mass NH4Cl = 365 grams

Explanation:

Step 1: Data given

Volume = 2.00 L

Molarity = 0.800M

pH = 8.63

Kb ammonia is 1.8*10^-5

Step 2: Calculate concentration NH4+

pH = pKa + log([A^-]/[HA])

Here, [A^-] = [NH3] (base)

[HA] = [NH4^+] (acid)

Rather than use Kb, it is easier to use the pKa of NH4^+ (the conjugate acid).

Kb = 1.8 * 10^-5  

pKb = -log(Kb) = -log(1.8 x 10^-5) = 4.74

pKa + pKb = 14

pKa = 14 - pKb = 14 - 4.74 = 9.26

pH = pKa + log([A^-]/[HA])

8.63 = 9.26 + log((0.800 M NH3)/(x M NH4^+))

-0.63 = log((0.800 M NH3)/(x M NH4^+))

(0.800 M NH3)/(x M NH4^+) = 10^-0.63 = 0.2344

x M NH4^+ = (0.800 M NH3 )/(0.2344)

x M NH4^+ = 3.413 M NH4^+

Step 3: Calculate moles NH4Cl

Moles NH4Cl = molarity * volumes

Moles NH4Cl = 3.413 M * 2.00 L

Moles NH4Cl = 6.826 moles NH4Cl

Step : Calculate mass NH4Cl

Mass NH4Cl = moles NH4Cl * molar mass NH4Cl

Mass NH4Cl = 6.826 moles * 53.49 g/mol

Mass NH4Cl = 365 grams

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.