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Drupady
2 months ago
10

In an acid-base neutralization reaction 43.74 ml of 0.500 m potassium hydroxide reacts with 50.00 ml of sulfuric acid solution.

what is the concentration of the h2so4 solution?'
Chemistry
2 answers:
lorasvet [2.7K]2 months ago
7 0
The neutralization reaction that occurs between potassium hydroxide and sulfuric acid can be represented as follows
2KOH + H2SO4 ---> K2SO4 + 2H2O

The quantity of moles of KOH is calculated using (43.74 x 0.500)/ 1000 = 0.02187 moles

Given that the stoichiometric ratio of KOH to H2SO4 is 2:1, the moles of H2SO4 can be determined as 0.02187/2 = 0.01094 moles

To find the concentration (molarity), use the formula (0.01094/50) x 1000 = 0.2188M
alisha [2.9K]2 months ago
3 0

Answer: The sulfuric acid's concentration is 0.219 M

Explanation:

We determine the acid concentration by applying the formula from the neutralization process:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 correspond to the n-factor, the molarity, and the volume of acid, whereas H_2SO_4

n_2,M_2\text{ and }V_2 denote the n-factor, molarity, and volume of the KOH base.

The provided data is:

n_1=2\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=43.74mL

Substituting these values into the aforementioned equation, we arrive at:

2\times M_1\times 50.00=1\times 0.500\times 43.74\\\\M_1=\frac{1\times 0.500\times 43.74}{2\times 50.00}=0.219M

Thus, the sulfuric acid concentration equates to 0.219 M

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2200 mg of antibiotic

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0.00004 multiplied by 55 (to find out the required amount for a 55 kg individual)

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Since Potassium (K) has an oxidation number of +1

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