What is the pOH concentration of a .0768 M NaOH solution?

1.2

Explanation:

pOH is the negative logarithm of hydroxide ion concentration in the solution.

Thus, pOH is expressed as

pH = - log(OH-)

NaOH(aq) ---> Na(aq) + OH-(aq)

The alkali ionizes to give hydroxide ion OH- as the only negative ion. Hence, the concentration of OH- is also 0.0768

So, pOH = - log(0 .0768)

pH = -(-1.15)

pH = 1.15 (since the two minus signs will cancel out)

Round the pOH value to the nearest tenth as 1.2

Thus, the pOH concentration of 0 .0768 NaOH solution is 1.2 (highly concentrated)

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o