A buffer solution contains 0.496 M KHCO3 and 0.340 M K2CO3. If 0.0585 moles of potassium hydroxide are added to 250. mL of this buffer, what is the pH of the resulting solution

pH = 10.9

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:

The resulting concentrations are:

Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:

Which makes sense since basic OH⁻ ions were added.

Regards!

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g