Enter a balanced equation for the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)

Explanation:

Let's consider the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.

Pb(NO₂)₂(aq) + LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)

This is a double displacement reaction. We will start balancing Cl by multiplying LiCl by 2.

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)

Now, we have to balance Li by multiplying LiNO₂ by 2.

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)

The equation is now balanced.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.