Enter the formula of the conjugate base for the acid HCO3

A.) Enter the formula of the conjugate base for the acid HCO3-. Express your answer as a chemical formula.

B.) Enter the formula of the conjugate base for the acid CH3-NH3+. Express your answer as a chemical formula.

C.) Enter the formula of the conjugate base for the acid HPO42-. Express your answer as a chemical formula.

D.) Enter the formula of the conjugate base for the acid HNO2. Express your answer as a chemical formula.

E.) Enter the formula of the conjugate base for the acid HBrO. Express your answer as a chemical formula.

The Correct Answer and Explanation is :

Answers:

A. ( \text{CO}_3^{2-} )
B. ( \text{CH}_3\text{-NH}_2 )
C. ( \text{PO}_4^{3-} )
D. ( \text{NO}_2^- )
E. ( \text{BrO}^- )


Explanation:

A conjugate base forms when an acid donates a proton (( \text{H}^+ )) during a chemical reaction. The conjugate base has one fewer proton and a corresponding change in charge compared to the acid.


A.) Conjugate base of ( \text{HCO}_3^- ):
The bicarbonate ion (( \text{HCO}_3^- )) can lose one proton to become ( \text{CO}_3^{2-} ) (carbonate ion). The reaction is:
[ \text{HCO}_3^- \rightarrow \text{CO}_3^{2-} + \text{H}^+ ]


B.) Conjugate base of ( \text{CH}_3\text{-NH}_3^+ ):
The methylammonium ion (( \text{CH}_3\text{-NH}_3^+ )) loses a proton to form ( \text{CH}_3\text{-NH}_2 ) (methylamine). The reaction is:
[ \text{CH}_3\text{-NH}_3^+ \rightarrow \text{CH}_3\text{-NH}_2 + \text{H}^+ ]


C.) Conjugate base of ( \text{HPO}_4^{2-} ):
The hydrogen phosphate ion (( \text{HPO}_4^{2-} )) loses a proton to become ( \text{PO}_4^{3-} ) (phosphate ion). The reaction is:
[ \text{HPO}_4^{2-} \rightarrow \text{PO}_4^{3-} + \text{H}^+ ]


D.) Conjugate base of ( \text{HNO}_2 ):
Nitrous acid (( \text{HNO}_2 )) loses a proton to form ( \text{NO}_2^- ) (nitrite ion). The reaction is:
[ \text{HNO}_2 \rightarrow \text{NO}_2^- + \text{H}^+ ]


E.) Conjugate base of ( \text{HBrO} ):
Hypobromous acid (( \text{HBrO} )) loses a proton to form ( \text{BrO}^- ) (hypobromite ion). The reaction is:
[ \text{HBrO} \rightarrow \text{BrO}^- + \text{H}^+ ]


General Rule:

  • A conjugate base is derived by removing one ( \text{H}^+ ) (proton) from the acid.
  • The charge decreases by 1 unit for each proton lost.
  • The chemical formula changes by removing the hydrogen atom.

This framework ensures accurate identification of conjugate bases.

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