If a pH of water is 5.3 then 4.3 is it 1.0 less acidic 1.0 less basic
2 answers:
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Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O
Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O
Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L
∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion =![\frac{number of moles of [OH-]}{volume of solution (L)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bnumber%20of%20moles%20of%20%5BOH-%5D%7D%7Bvolume%20of%20solution%20%28L%29%7D%20)
= 
= 0.03199 M
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O
Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O
Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L
∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion =
Answer:
1255.4L
Explanation:
Given parameters:
P₁ = 928kpa
T₁ = 129°C
V₁ = 569L
P₂ = 319kpa
T₂ = 32°C
Unknown:
V₂ = ?
Solution:
The combined gas law application to this problem can help us solve it. It is mathematically expressed as;
P, V and T are pressure, volume and temperature
where 1 and 2 are initial and final states.
Now,
take the units to the appropriate ones;
kpa to atm, °C to K
P₂ = 319kpa in atm gives 3.15atm
P₁ = 928kpa gives 9.16atm
T₂ = 32°C gives 273 + 32 = 305K
T₁ = 129°C gives 129 + 273 = 402K
Input the values in the equation and solve for V₂;
V₂ = 1255.4L