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(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.

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jufsanabriasa

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

annyksl

The sample would have 26.20%w/w of KBr.

We can arrive at this answer as follows:

  • First, we have to determine the number of liters in the solution, for that, use the following calculation:

[tex]25.13mL-0.65mL = 24.48mL\\\\\\{{1000mL=1L} \atop {24.48mL=X}} \right. \\1000x = 24.48\\x= \frac{24.48}{1000} = 0.02448L[/tex]

  • From this, we can calculate the number of moles of AgNO₃ in the solution, which will be equal to the number of moles of KBr. This calculation will be done as follows:

[tex]0.02448L* (0.04614mol/L) = 1.130x10^-^3[/tex] moles of AgNO₃

  • With this value, we should calculate the molar mass of KBr. This calculation will be done as follows:

[tex]1.130x10^-^3* (119g / 1mol) = 0.1344g[/tex]

  • We now have all the values needed to calculate the %w/w in the sample. This will be done as follows:

[tex]\frac{0.1344}{(0.5131*100)} =26.20[/tex] % w/w

With that, we can say that the sample will have 26.20% w/w of KBr.

More information:

https://brainly.com/question/7944314?referrer=searchResults

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