274 mL H3 O+ and fully neutralized
It will take one teaspoon of Mg(OH)2 to completely neutralize 2.00×10^2mL of H3O+.
<h3>What is the purpose of milk of magnesia?</h3>
- For a brief period of time, this medicine is used to relieve sporadic constipation.
- It is an osmotic laxative, which means that it works by drawing water into the intestines, which aids in causing bowel movement.
<h3>What dosage of milk of magnesia is recommended for constipation?</h3>
- Take Milk of Magnesia once day, preferably before bed, in divided doses, or as prescribed by a physician.
- suggested dosage: 30 mL to 60 mL for adults and kids 12 years of age and older. 15 mL to 30 mL for children aged 6 to 11 years.
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the question you are looking for is
People often take milk of magnesia to reduce the discomfort associated with acid stomach or heartburn. The recommended dose is 1 teaspoon, which contains 4.00x 10^{2} mg of Mg(OH)_2. What volume of an HCl solution with a pH of 1.3 can be neutralized by one dose of milk of magnesia? If the stomach contains 2.00x10^{2}mL of pH 1.3 solution, is all the acid neutralized? If not, what fraction is neutralized?
Answer:
a. True
Explanation:
Theoretical yield is the amount of product that could be obtained if a chemical reaction has 100% efficiency.
Hope it helps...
Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:
Whereas the moles of the salt are computed as shown below:
So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
![[salt]=0.01428mol/0.0276L=0.517M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D0.01428mol%2F0.0276L%3D0.517M)
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:
Whose equilibrium expression is:
![Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5NH_2%5D%5BH_3O%5E%2B%5D%7D%7BC_6H_5NH_3%5E%2B%7D)
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:
Whereas x is:
Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:
Regards!
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