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AleksandrR [38]
4 years ago
10

4 physical properties of spoonfull of sugar

Chemistry
2 answers:
kirill [66]4 years ago
6 0
Um, shiny, metal, grainy, white, silver, small
almond37 [142]4 years ago
3 0
Small crystals 
<span>white, brown </span>
<span>hard as in solid at room temp </span>
<span>sweet </span>
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Read 2 more answers
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is t
Artist 52 [7]

Answer:

pKa = 3.51

Explanation:

The titration of acid solution with NaOH can be illustrated as:

AH + NaOH \to NaA + H_2O

Given that:

Volume of acid solution (V_1) = 25 \ mL

Volume of NaOH (V_2) = 18.8 \ mL

Molarity of acid solution (M_1) = ???

Molarity of NaOH (M_2) = 0.10 \ M

For Neutralization reaction:

M_1V_1 = M_2V_2

Making M_1 the subject of the formula; we have:

M_1 = \frac{M_2V_2}{V_1}

M_1 = \frac{0.10*18.8}{25}

M_1 =0.0752 \ M  

However; since the number of moles of NaA formed is equal to the number of moles of NaOH used : Then :

M_2V_2 = 0.10 *18.8 = 1.88 \ mm

Total Volume after titration = ( 25 + 18.8 ) m

= 43.8 mL

Molarity of salt (NaA ) solution = \frac{number \ of \ moles}{Volume \  (mL)}

= \frac{1.88}{43.8}

= 0.0429 M

After mixing the two solution ; the volume of half neutralize solution is = 25 mL + 43.8 mL

= 68.8 mL

Molarity of NaA before mixing M_1 = 0.0429 \ M

Volume (V_1) = 43.8 \ mL

Molarity of NaA after mixing M_2 = ???

Volume (V_2 ) = 68.8 \ mL

∴

M_2 = \frac{M_1*V_1}{V_2} \\ \\ M_2 = \frac{0.0429*43.1}{68.8} \\ \\ M_2 = 0.0273 \ M

Molarity of acid before mixing = 0.0725 M

Volume = 25 mL

Molarity of acid after mixing = \frac{0.0752*25}{68.8}

= 0.0273 M

Since this is a buffer solution ; then using Henderson Hasselbalch Equation

pH = pKa + log \frac{[salt]}{[acid]}

3.51= pKa + log \frac{[0.0273]}{[0.0273]} \\  \\ 3.51= pKa + log \ 1  \\ \\ 3.51= pKa + 0 \\ \\ pKa = 3.51

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