Chemistry: Molecular Approach (4th Edition)

1. Calculate the molar mass of HI: $HI$ : ( 1.008 $\times$ 1 )+ ( 126.9 $\times$ 1 )= 127.9 g/mol a. $$[H_3O^+] = 10^{-pH} = 10^{-1.25} = 5.6 \times 10^{-2} \space M$$ Since HI is a strong acid, $[HI] = 5.6 \times 10^{-2} \space M$ $$0.250 \space L \times \frac{5.6 \times 10^{-2} \space mol}{ 1 \space L} \times \frac{127.9 \space g}{1 \space mol}= 1.8 \space g$$ b. $$[H_3O^+] = 10^{-pH} = 10^{-1.75} = 1.8 \times 10^{-2} \space M$$ Since HI is a strong acid, $[HI] = 1.8 \times 10^{-2} \space M$ $$0.250 \space L \times \frac{1.8 \times 10^{-2} \space mol}{ 1 \space L} \times \frac{127.9 \space g}{1 \space mol}= 0.57 \space g$$ c. $$[H_3O^+] = 10^{-pH} = 10^{-2.85} = 1.4 \times 10^{-3} \space M$$ Since HI is a strong acid, $[HI] = 1.4 \times 10^{-3} \space M$ $$0.250 \space L \times \frac{1.4 \times 10^{-3} \space mol}{ 1 \space L} \times \frac{127.9 \space g}{1 \space mol}= 0.045 \space g$$