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cluponka [151]
2 years ago
10

Twenty-seven milliliters of an acid with an unknown concentration are titrated with a base that has a concentration of 0.55 M. T

he indicator changed color when 12.5 milliliters of base were added. What is the concentration of the unknown acid? 0.25 M 1.2 M
Chemistry
2 answers:
murzikaleks [220]2 years ago
5 0

Twenty-seven milliliters of an acid with an unknown concentration are titrated with a base that has a concentration of 0.55 M. The indicator changed color when 12.5 milliliters of base were added. The concentration of the unknown acid is 0.25 M.

Svetllana [295]2 years ago
4 0
It should be 0.25 M. Use the formula C1*V1=C2*V2, for those values, as it is right when it changed colour. Remember to change the if those are not the same (but in your case it is, so no need this time). 

C1*V1=C2*V2
C1*27ml=0.55M*12.5ml
C1=(0.55M*12.5ml)/27ml = 0.25M
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Answer:

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Q_{sample} - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

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Where:

m_{s}, m_{w} - Mass of water and the sample of aluminium, measured in kilograms.

c_{p,s}, c_{p,w} - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

T_{s}, T_{w} - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

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The final temperature is now cleared:

(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}

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T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}

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