Consider the titration of 20.00 ML of KOH 0.01 mol/L with HNO3 0.01 mol/L. Calculate the PH of the titration solution after the addition of the following titration volumes:
A. 0.00 ml C. 20.00 ml B. 19.99 ml D. 25.00 ml
You can do 0.00 mL can't you? That's just 0.01 M KOH solution. That gives you the OH and you convert to pH.
The 20.00 mL addition SHOULD tell you that is the equivalence point which means you have pure NaCl at the point in water. What do you think the pH is when you know that neither Na%+ nor Cl^- are hydrolyzed.
For the 19.99 mL. I'll do millimols to keep the number of zeros down..
millimols KOH = 20.00 x 0.01 = 0.2000
millimols HNO3 = mL x M = 19.99 x 0.01 = 0.1999
Excess KOH = 0.2000-0.1999 = ?
M KOH excess = millimols/mL = ?
Convert KOH M to OH (or pOH) and then to pH.
thank you so much :)))
To calculate the pH of the titration solution after the addition of different titration volumes, we need to understand the reaction between KOH and HNO3 and how it affects the pH.
First, let's write the balanced chemical equation for the reaction between KOH and HNO3:
KOH + HNO3 -> KNO3 + H2O
From the equation, we can see that the reaction between KOH and HNO3 results in the formation of water (H2O) and potassium nitrate (KNO3).
Since both KOH and HNO3 are strong electrolytes, they will dissociate completely in water. KOH will dissociate into K+ ions and OH- ions, while HNO3 will dissociate into H+ ions (protons) and NO3- ions.
Now, let's examine the effect of the titration volumes on the pH:
A. 0.00 mL:
In this case, no HNO3 has been added, so there will be no reaction, and the solution will be purely KOH. Since KOH is a strong base, it will dissociate completely into K+ ions and OH- ions. The concentration of OH- ions will be determined by the initial concentration of KOH (0.01 mol/L) and the volume of the solution (20.00 mL). With this information, you can calculate the pOH of the solution using the formula: pOH = -log[OH-]. To convert pOH to pH, you can use the relationship: pH + pOH = 14.
B. 19.99 mL:
At this point, almost all of the HNO3 has reacted with the KOH, resulting in a mostly neutral solution. The OH- ions from KOH will react with the H+ ions from HNO3, forming water. However, there will be a small excess of OH- ions remaining. To calculate the pH at this point, you need to determine the concentration of OH- ions using the remaining amount of KOH and water volume (20.00 mL - 19.99 mL).
C. 20.00 mL:
At this point, all of the HNO3 has reacted with the KOH, resulting in the complete neutralization of the H+ ions and the OH- ions. The solution will contain only water (H2O) and potassium nitrate (KNO3). Since KNO3 is a neutral salt, it will not affect the pH of the solution. Hence, the pH will be determined purely by the autoionization of water, which is pH 7.
D. 25.00 mL:
In this case, an excess volume of HNO3 has been added, resulting in the solution being acidic. The excess H+ ions will determine the pH of the solution. To calculate the pH, you need to consider the concentration of H+ ions from the excess HNO3 and the remaining volume of the solution (25.00 mL - 20.00 mL).
By following these steps and considering the chemical reaction and stoichiometry involved, you can calculate the pH for each specific titration volume.