question
when bases are introduced into the blood, the concentration of $oh^{-}$ is increased. the bicarbonate buffer system acts to neutralize this increase through the reaction show below. what is the missing product?
$oh^{-}+h_{2}co_{3}
ightleftharpoons h_{2}o+$______
• write the missing product as a chemical formula along with any charge.
• do not include the state (phase).
provide your answer below:

Question

Accepted Answer

# Explanation:
## Step1: Analyze the reaction
We know that in a chemical reaction, atoms are conserved. Let's count the atoms on both sides.
On the left - hand side:
- For $H$: In $OH^-$, there is $1$ $H$ atom, and in $H_2CO_3$, there are $2$ $H$ atoms. So in total, there are $1 + 2=3$ $H$ atoms. In $H_2O$ (on the right - hand side), there are $2$ $H$ atoms.
- For $C$: There is $1$ $C$ atom in $H_2CO_3$.
- For $O$: In $OH^-$, there is $1$ $O$ atom, and in $H_2CO_3$, there are $3$ $O$ atoms. So in total, there are $1+3 = 4$ $O$ atoms. In $H_2O$, there is $1$ $O$ atom.

Also, consider the charge. The left - hand side has a net charge of $- 1$ (from $OH^-$, since $H_2CO_3$ is neutral).

## Step2: Determine the product
Let the product be $X$.
We know that $OH^-+H_2CO_3 = H_2O+X$
By conservation of atoms and charge:
The chemical formula of $X$ should be $HCO_3^-$
 - For $H$: In $H_2O$ ($2$ $H$ atoms) and $HCO_3^-$ ($1$ $H$ atom), total $2 + 1=3$ $H$ atoms (matches the left - hand side).
 - For $C$: $1$ $C$ atom (matches the left - hand side).
 - For $O$: In $H_2O$ ($1$ $O$ atom) and $HCO_3^-$ ($3$ $O$ atoms), total $1+3 = 4$ $O$ atoms (matches the left - hand side).
 - For charge: $HCO_3^-$ has a charge of $-1$ (matches the net charge of the left - hand side, since $H_2CO_3$ is neutral and $OH^-$ has a charge of $-1$)

# Answer:
$HCO_3^-$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Analyze the reaction

Step2: Determine the product

Answer: