Question

Sodium rhodizonate is often used in forensic investigations to detect the presence of any residue from shooting a firearm. Burned and unburned particles from the gunpowder will be ejected with the projectile. In the photograph below, the residue from the gunshot is stained red by sodium rhodizonate at a pH of 2.8. How many mL of 0.65 M hydrochloric acid need to be added to a neutral solution of sodium rhodizinate to make 500.0 mL of solution with a pH of 2.800?

Answer 1

Question #78733

Sodium rhodizonate is often used in forensic investigations to detect the presence of any residue from shooting a firearm. Burned and unburned particles from the gunpowder will be ejected with the projectile. In the photograph below, the residue from the gunshot is stained red by sodium rhodizonate at a pH of 2.8. How many mL of 0.65 M hydrochloric acid need to be added to a neutral solution of sodium rhodizinate to make 500.0 mL of solution with a pH of 2.800?

Answer:

The precise formula for calculating the pH and pOH is

$$\mathrm{pH} = - \log [\mathrm{H}^{+}]$$ $$\mathrm{pOH} = - \log [\mathrm{OH}^{-}]$$ $$[\mathrm{H}^{+}] = 10^{-\mathrm{pH}}$$ $$[\mathrm{OH}^{-}] = 10^{-\mathrm{pOH}}$$

Therefore:

$$[\mathrm{H}^{+}] = 10^{-\mathrm{pH}}$$ $$[\mathrm{H}^{+}] = 10^{-2.8} = 0.001584$$

And

$$\mathrm{V} = 500 \mathrm{ml} = 0.5 \mathrm{l}$$ $$\mathrm{C_M} = \mathrm{n}/\mathrm{V} \quad \mathrm{n} = \mathrm{C_M} \times \mathrm{V} \quad \mathrm{n} = 0.001584 \times 0.5 = 0.0008 \mathrm{mol}$$ $$\mathrm{V} = \mathrm{n}/\mathrm{C_M} = 0.0008 / 0.65 = 0.00123 \mathrm{l}$$ $$\mathrm{V} = 0.00123 \mathrm{l} = 1.23 \mathrm{ml}.$$

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