Determine the ph of a 0.227 m c5h5n solution at 25°c. the kb of c5h5n is 1.7 × 10-9.

 the concentration of OH- using the formula:
[OH-] = Kw / [H+]

[OH-] = 1.0 x 10^-14 / ( [(1.7 x 10^-9) (0.227)]^2 )

[OH-] = 5.1 x 10^-10

To find the pOH, use the formula:
pOH = -log [OH-]

pOH = -log [5.1 x 10^-10]
pOH = 9.29

Explanation:2

Na

(

s

)

+

Cl

2

(

g

)

→

2

NaCl

(

s

)

and that the reaction produced  

234 g

of sodium chloride. Convert this to moles by using the compound's molar mass

234

g

⋅

1 mole NaCl

58.44

g

=

4.004 moles NaCl

Now, you know that the sample of sodium metal reacted completely, which implies that it was completely consumed by the reaction.

In other words, you don't have to worry about the sample of chlorine gas because the fact that sample of sodium metal was completely consumed lets you know that the chlorine gas is not a limiting reagent.

This means that the reaction consumed

4.004

moles NaCl

⋅

2 moles Na

2

moles NaCl

=

4.004 moles Na

Convert this to grams by using the element's molar mass

4.004

moles Na

⋅

22.99 g

1

mole Na

=

92.1 g

−−−−−

 

The answer is rounded to three sig figs.

SIDE NOTE You can show that chlorine gas is not the limiting reagent by calculating the number of moles present in the sample

0.142

kg

⋅

1 mole Cl

2

35.453

kg

=

4.005 moles Cl

2

In order to produce  

4.004

moles of sodium chloride, you only need

4.004

moles NaCl

⋅

1 mole Cl

2

2

moles NaCl

=

2.002 moles Cl

2

Since you have more chlorine gas than you need, chlorine is in excess.