## Chemistry: Molecular Approach (4th Edition)

a. 1. Since $HI$ is a strong acid, all its concentration is going to produce $H_3O^+$: $$[H_3O^+] = [ HI ] = 0.048 \space M$$ $$pH = -log[H_3O^+] = -log( 0.048 ) = 1.32$$ b. 1. Since $HClO_4$ is a strong acid, all its concentration is going to produce $H_3O^+$: $$[H_3O^+] = [ HClO_4 ] = 0.0895 \space M$$ $$pH = -log[H_3O^+] = -log( 0.0895 ) = 1.048$$ c. 1. Since $HClO_4$ and $HCl$ are both strong acids, all their concentrations are going to produce $H_3O^+$; we just have to sum the total concentration: $$[H_3O^+] = 0.045 + 0.048 = 0.093 \space M$$ $$pH = -log[H_3O^+] = -log( 0.093 ) = 1.03$$ d. 1. Calculate the molar mass: $HCl$ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol 2. Use the information as conversion factors to find the molarity of this solution: $$\frac{ 1.09 g \space HCl }{100 \space g \space solution} \times \frac{1 \space mol \space HCl }{ 36.46 \space g \space HCl } \times \frac{ 1.01 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 0.302 \space M$$ 3. Since $HCl$ is a strong acid, all its concentration is going to produce $H_3O^+$: $$[H_3O^+] = [ HCl ] = 0.302 \space M$$ $$pH = -log[H_3O^+] = -log( 0.302 ) = 0.520$$