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rusak2 [61]
3 years ago
7

In a titration experiment, 31.4 mL of 1.120 M HCOOH is neutralized by 16.3 mL of Ba(OH)2. What is the concentration of the Ba(OH

)2 solution?
Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0

Answer : The concentration of the Ba(OH)_2 solution is, 2.16 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCOOH

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ba(OH)_2.

We are given:

n_1=1\\M_1=1.120M\\V_1=31.4mL\\n_2=2\\M_2=?\\V_2=16.3mL

Putting values in above equation, we get:

1\times 1.120M\times 31.4mL=2\times M_2\times 16.3mL\\\\M_2=2.16M

Thus, the concentration of the Ba(OH)_2 solution is, 2.16 M

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Answer: 40.3 L

Explanation:

To calculate the moles :

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Calculate the concentration (in mol/l) of 33% by weight (33 g naoh per 100 g of solution) naoh solution. (the density of the 33%
Shtirlitz [24]
Molarity is measured in moles per Liter. If there are 1.35 g/mL, find out how many grams there are in a liter of solution.

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By weight, the NaOH is 33% or .33

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