Chemistry Notes: Determination of Equivalent Weight of Monoprotic Acid PDF

### Determination of Equivalent Weight of Monoprotic Acid * **Arrhenius 1887** * **Acid:** A substance that dissociates in water to increase the concentration of $H^+$. * **Base:** A substance that dissociates in water to increase the concentration of $OH^-$. * **Brønsted and Lowry** *...

### Determination of Equivalent Weight of Monoprotic Acid * **Arrhenius 1887** * **Acid:** A substance that dissociates in water to increase the concentration of $H^+$. * **Base:** A substance that dissociates in water to increase the concentration of $OH^-$. * **Brønsted and Lowry** * **Acid:** A substance which has a tendency to donate a proton. * **Base:** A substance which has a tendency to accept a proton. * Example: $HCl + H_2O \rightarrow H_3O^+ +Cl^-$, where $H_3O^+$ is the hydronium ion. * **Lewis** * **Acid:** A substance which accepts a pair of electrons in reaction. * **Base:** A substance which donates a pair of electrons in reaction. * Example: $F:B:O + :N:H \rightarrow F:B:H:N:H$ where boron is the acid (accept) and nitrogen is the base (donate). ### Types of Acids * **Monoprotic Acids:** * $HCl$, $HNO_3$, $CH_3COOH$, $HCOOH$ * **Polyprotic Acids:** * $H_2SO_4$, $H_3PO_4$ * $H_2SO_4 \rightarrow H^+ + HSO_4^-$ * $HSO_4^- \rightarrow H^+ + SO_4^{-2}$ * $H_3PO_4 \rightarrow H^+ + H_2PO_4^{-}$ * $H_2PO_4^- \rightarrow H^+ + HPO_4^{-2}$ * $HPO_4^{-2} \rightarrow H^+ + PO_4^{-3}$ **Note:** The Arabic text in the document is not translated. ### Molarity Molarity is defined as the *number of moles of solute per 1 litre of solution*. $M = \frac{No. of moles of solute}{Volume}$ $No. of moles of solute = \frac{weight}{Molecular weight}$ ### Normality Normality is defined as the *number of gram equivalents of solute per 1 litre of solution*. $N = \frac{no. of gram equivalents}{Volume}$ $no. of gram equivalents = \frac{wt}{equivalent weight} = \frac{wt}{eq.wt}$ $equivalent weight = \frac{Molecular weight}{no. of H^+}$ (for acids) $equivalent weight = \frac{Molecular weight}{no. of (OH)^-}$ (for bases) $equivalent weight = \frac{Molecular weight}{عدد أحد الشقوق × التكافؤ}$ (for salts)