(I think you have a mistake in your question as the addition is 30mL, not 100mL)
when PH = - ㏒[H+]
and here we have HClO4 is the strong acid
So PH = - ㏒[HClO4]
moles of HClO4 = 0.1 L *0.18 m = 0.018 M
moles of LiOH = 0.03 L * 0.27 m = 0.0081 M
when the total volume = 0.1L + 0.03L = 0.13 L
∴ [HClO4] = (0.018-0.0081)/0.13 L
= 0.076 M
PH = -㏒ 0.076
= 1.12
The correct answer and the proper form for this question to be answered could be found on google
The answer is 546K. If you need the answer in degrees Celsius you subtract 273 from 546 and get 273 degrees Celsius. I included how I got the answer in the photo below. Hope this helps!
When titrating a strong monoprotic acid and KOH at 25 degrees Celcius, the pH will be approximately equal to 7.00. Since a strong acid and a strong base reacts through a neutralization reaction. The pH of a neutral solution is 7.00. <span />
Answer:
initial temperature=
Explanation:
Assuming that the given follows the ideal gas nature;
mole of gass will remain same at any emperature:
putting all the value we get:
initial temperature=