Category: Chemical Engineering Math
"Published in Newark, California, USA"
In standardizing a solution of NaOH against 1.431 grams of potassium hydrogen phthalate, C8H5KO4 (KHP), the analyst uses 35.50 mL of the alkali and has to run back with 8.25 mL of acid (1 mL = 10.75 mg NaOH). What is the molarity of the NaOH solution?
Solution:
The given word problem is about volumetric analysis problem because it involve the titration of samples and solutions. Potassium hydrogen phthalate (KHP) is a sample used to standardize the basic or alkali solutions that are prepared in the lab.
If NaOH solution is used to titrate KHP sample, then the chemical reaction for neutralization is
The concentration of NaOH solution used to titrate a sample of KHP is
Since, there's a back titration with an acid solution especially if there's an over titration with an alkali solution, then the above equation becomes
Usually, we put the indicator at the sample solution so that we know that we used the right amount of acid or alkali solution by its color. The color of the indicator is changed once it reached its end point. How about if you used the excess solution for titration, then we have to do the back titration until we get the correct color for its end point.
Therefore, the concentration of NaOH solution used to titrate a sample of KHP is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
Calculate the molarity of NaOH solution if 12.25 mL was used to titrate 0.2615 gram of primary standard KHP (potassium hydrogen phthalate, C8H5KO4).
Solution:
The given problem is about finding the concentration of NaOH solution. KHP is an acid mostly used for standardization of any basic solution that are usually prepared in the lab. Although we know how to prepare any solutions in the lab, it is better to do the standardization so that we can check if the concentration is correct or not.
If NaOH solution is used to titrate KHP sample, then the chemical reaction for neutralization is
Therefore, the concentration of NaOH solution used to titrate a sample of KHP is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
What is the pH of the resulting solution made by mixing 5 mL of 0.2178 M HCl and 15 mL of 0.1156 M NH4OH? Kb = 1.8 x 10-5 for NH4OH.
Solution:
If an acid is mixed with a base, then the resulting product is a salt. If you mix a strong acid with a strong base, then the resulting product is a neutral salt. If you mix a weak acid with a weak base, then the resulting product is a neutral salt. However, if you mix a strong acid with a weak base, then the resulting product is an acidic salt. If you mix a weak acid with a strong base, then the resulting product is a basic salt.
The pH of any neutral salt is usually equal to 7. The pH of an acidic salt is less than 7 and the pH of a basic salt is greater than 7.
In the given problem, if HCl is mixed with NH4OH, then the resulting product is
Since NH4Cl is an acidic salt, then the pH is less than 7. From the given amount and concentration of the reactants, let's see if the resulting solution is acidic or basic as follows
Since the number of moles of NH4OH is greater than with the number of moles of HCl, then the resulting mixture is basic. Well, let's see.
Hence, the resulting concentration of NH4OH is
Since NH4OH is a weak base, then its ionization is written as follows
The concentration of [OH-] is
Since the value of Kb is very small, then we can neglect x at the denominator as follows
The pH of NH4OH is
The resulting concentration of NH4Cl which is their product is
The ionization of NH4Cl is written as follows
Since [H+] is the product of the ionization process, then we need to convert Kb into Ka as follows
The concentration of [H+] in NH4Cl is
Since the value of Ka is very small, then we can neglect x at the denominator as follows
The pH of NH4Cl is
Therefore, the pH of the overall resulting solution is
