Explanation:
A.
In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.
B.
Equation of the reaction
2NaOH + H2SO4 --> Na2SO4 + 2H2O
Number of moles of H2SO4 = molar concentration × volume
= 0.75 × 0.0105
= 0.007875 moles.
By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875
= 0.01575 moles.
Molar concentration of NaOH = number of moles ÷ volume
= 0.01575 ÷ 0.0175
= 0.9 M of NaOH.
Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
1.Start with the number of grams of each element, given in the problem.
2.Convert the mass of each element to moles using the molar mass from the periodic table.
3.Divide each mole value by the smallest number of moles calculated.
4.Round to the nearest whole number. This is the mole ratio of the elements and is.
760.0 mmHg(V)=600.0mmHg(240.0mL)
600(240)/760=189mL
