Question

(Direct Titration) A 0.5131-g sample that contains KBr is dissolved in 50 mL of distilled water. Titrating with 0.04614 M AgNO3 requires 25.13 mL to reach the Mohr end point. A blank titration requires 0.65 mL to reach the same end point. Report the %w/w KBr in the sample.

Answer 1

Answer:

26.20% w/w of KBr in the sample

Explanation:

Mohr titration is a way to quantify Br⁻ and Cl⁻ ions in solution. The reaction is:

KBr(aq) + AgNO₃(aq) → KNO₃(aq) + AgBr(s)

where 1 mole of KBr reacts per mole of AgNO₃

Thus, moles of AgNO₃ in (25.13mL-0.65mL = 24.48mL) of a 0.04614M solution are:

0.02448L × (0.04614mol / L) = 1.130x10⁻³ moles of AgNO₃ = moles of KBr.

Mass of KBr -Molar mass: 119g/mol- is:

1.130x10⁻³ moles of KBr × (119g / 1mol) = 0.1344g of KBr

Thus, %w/w of KBr in the sample is:

%w/w = 0.1344g / 0.5131g ×100 = 26.20% w/w of KBr in the sample