Answer
a. The mass of HI is equal to 1.8 g
b. The mass of HI is equal to 0.57 g
c. The mass of HI is equal to 0.045 g
Work Step by Step
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, $[HI] = 1.4 \times 10^{-3} \space M$
$$0.250 \space L \times \frac{1.4 \times 10^{-3} \space mol}{ 1 \space L} \times \frac{127.9 \space g}{1 \space mol}= 0.045 \space g$$
