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lawyer [7]
3 years ago
5

A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th

e solution. Round your answer to decimal places.
Chemistry
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

pH=4.1

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1

Regards.

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3 years ago
Can you please help me and can you show your work please
natali 33 [55]

Answer:

1. 25 moles water.

2. 41.2 grams of sodium hydroxide.

3. 0.25 grams of sugar.

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Explanation:

Hello!

In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:

1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:

1.5x10^{25}molecules*\frac{1mol}{6.022x10^{23}molecules} =25 molH_2O

2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:

1.2mol*\frac{40.00g}{1mol} =41.2g

3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:

45g*\frac{1mol}{180.15g}=0.25g

4. Here, since the molar mass of ammonia is 17.03 g/mol:

20mol*\frac{17.03g}{1mol}=340.6g

5. Here, since the molar mass of SO2 is 64.06 g/mol:

48g*\frac{1mol}{64.06g} *\frac{6.022x10^{23}molecules}{1mol} =4.5x10^{23}molecules

Best regards!

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3 years ago
Which of the following is TRUE?Which of the following is TRUE?A basic solution does not contain H3O+An neutral solution does not
schepotkina [342]

Answer:

b

Explanation:

b

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