Answer:
![p[H+] = 10.042](https://tex.z-dn.net/?f=p%5BH%2B%5D%20%3D%2010.042)
Explanation:
As we know that
......eq (1)
we will calculate the pH of OH- and then we will calculate the pH of H+
So p[OH-] ![= - log [1.10 * 10^{-4}]](https://tex.z-dn.net/?f=%3D%20-%20log%20%5B1.10%20%2A%2010%5E%7B-4%7D%5D)
Solving the right side of the equation, we get
p[OH-]
![= - [-3.958]\\= 3.958](https://tex.z-dn.net/?f=%3D%20-%20%5B-3.958%5D%5C%5C%3D%203.958)
Now we know that
Substituting the value of pOH in the above equation, we get -
![14.0 = p[H+] + 3.958\\p[H+] = 14 - 3.958\\p[H+] = 10.042](https://tex.z-dn.net/?f=14.0%20%3D%20p%5BH%2B%5D%20%2B%203.958%5C%5Cp%5BH%2B%5D%20%3D%2014%20-%203.958%5C%5Cp%5BH%2B%5D%20%3D%2010.042)
Answer:
yes it gives some bad effect
Explanation:
M ( HCl ) = ?
V ( HCl ) = 25.5 mL in liters : 25.5 / 1000 => 0.0255 L
M ( NaOH ) = 0.113 M
V ( NaOH ) = 51.2 mL / 1000 => 0.0512 L
number of moles NaOH:
n = M x V
n = 0.113 x <span> 0.0512 => 0.0057856 moles of NaOH
mole ratio:
</span><span>HCl + NaOH = NaCl + H2O
</span><span>
1 mole HCl -------------- 1 mole NaOH
( moles HCl ) ----------- </span><span> 0.0057856 moles NaOH
</span>
(moles HCl ) = <span> 0.0057856 x 1 / 1
</span>
= <span> 0.0057856 moles of HCl
</span>
M ( HCl ) = n / V
M = 0.0057856 / <span>0.0255
</span>
= 0.227 M
Answer A
hope this helps!
