0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.
The balanced neutralization equation is:
NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)
- Step 1: Calculate the reacting moles of KHP.
0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.
0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol
- Step 2: Determine the reacting moles of NaOH.
The molar ratio of NaOH to KHP is 1:1.
1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH
- Step 3: Calculate the molarity of NaOH.
1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.
[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Learn more about titration here: brainly.com/question/4225093
10 miles per hour because 500 divided by 50 is 10
Ionization energy is directly proportional to elements location on the periodic table
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:
Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH)
With the values put in: 0.35 x 45 = 0.35 x V(NaOH)
= 45 ml.
There is 45 ml of V(NaOH)
Let me know if you need anything else. :)
- Dotz
Correct answer is option E. <span>It is a redox reaction in which Zn is oxidized at the anode, and V is reduced at the cathode.
Reason:
In above reaction, the oxidation state of VO3- is +5, while that of VO2 is +4. Thus there is reduction of V from +5 to +4
In case of Zn, oxidation state of Zn is increased from 0 to +2, Thus process is referred as oxidation. </span>