What is the ph of a solution that has a hydroxide ion concentration of 1.0×10−9 m?

Explanation:

Step 1. Calculate the pOH

pOH =-log[OH⁻]

pOH =-log(1.0 × 10⁻⁹)

pOH = 9.00

Step 2. Calculate the pH

pH + pOH = 14.00

pH + 9.00 = 14.00

first one is the correct answer

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