need help with these ques thanks
For the first question, the first thing you need to do is find the moles of NaOH, which I can see you have already done. This will equal the moles of OH
-, and thus the moles of H
+ used in neutralisation. To find the moles of citric acid, we first need to consider the proticity of the acid. As the acid is triprotic, this value will be a
third of the moles of H
+. Once you have the molar value for the acid, you can find the concentration of it by using the given volume. There is a dilution factor of 10, thus you can obtain the original concentration of the undiluted acid by multiplying the concentration you get by 10
This question provides way more information than is required to answer it!
For the second question, the first step is to find the moles of H
+ remaining after neutralisation - this is where you got up to. From here, what you need to do is find [H
+] for the final volume of the solution - 100mL. Then you can find the pH of solution (pH=-log[H
+]), which works out to be B
how to answer this ques:
Using indicator paper, 0.1 mol/L solutions of NaCH3COO and NH4Cl are found to have pH values of 9 and 5, respectively. NaCl solution is neutral. Account for the observed pH values of NaCH3COO and NH4Cl in solutions. 3 marks
As the concentration of both solutions is the same, and there is a variation in pH, what you will need to discuss is the relative
strengths of each of the solutions. To do this, it is easiest to observe each compound's behaviour when dissolved in water.
Sodium acetate(NaCH
3COO), when dissolved, dissociates into sodium(Na
+) and acetate (CH
3COO
-) ions.
}\longrightarrow\text{Na}^{+}_{(aq)}+\text{CH}_{3}\text{COO}^{-}_{(aq)})
As we are aware that NaCl solution is neutral, the Na
+ ion can be ignored. When reacted with water, the acetate ion forms acetic acid (CH
3COOH) and a free hydroxide ion (OH
-)
}+\text{H}_{2}\text{O}_{(l)}\longrightleftharpoons\text{CH}_{3}\text{COOH}_{(aq)}+\text{OH}^{-}_{(aq)})
Due to the free hydroxide ion, and the weak nature of acetic acid(as it only partly dissociates in water), it is safe to assume that the solution will be basic. (pH = 9)
Ammonium chloride (NH
4Cl), when dissolved, dissociates into ammonium(NH
4+) and chloride(Cl
-) ions.
}\longrightarrow\text{NH}^{+}_{4(aq)}+\text{Cl}^{-}_{(aq)})
Like earlier, as we are aware that NaCl solution is neutral, the Cl
- ion can be ignored. When reacted with water, the ammonium ion forms ammonia(NH
3) and a hydronium(H
3O
+) ion.
}^{+}+\text{H}_{2}\text{O}_{(l)}\longrightleftharpoons\text{NH}_{3(aq)}+\text{H}_{3}\text{O}^{+}_{(aq)})
Due to the free hydronium ion, and the weak nature of ammonia as a base(as it only partly dissociates in water), it is safe to assume that the solution will be acidic. (pH = 5)
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