It isn't balanced. You have 3 *H and 3 * O, so something in the first formula must be changed. Have been searching for 1 hour and can't find the answer.
The pH of the weak acid is 6.93.
Given,
K= 1.15 *
c=0.102
[H]= K*c= 1.15 *
* 0.102= 0.1173 *
p[H]= -log[H]= 6.93
<h3>Weak acid</h3>
A weak acid is one that partially separates into its ions in water or an aqueous solution. A strong acid, on the other hand, completely splits into its ions in water. The conjugate base of a weak acid is also a weak base, and vice versa for the conjugate acid of a weak base. Weak acids have a higher pH than strong acids at the same concentration. Simple arrows pointing left to right are used to represent the reaction of a strong acid ionising in water. On the other hand, a weak acid ionising in water has a double arrow reaction arrow, indicating that both the forward and reverse reactions take place at equilibrium.
Learn more about Weak acid here:
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Did you intend to write [PdCl4]^-2 instead of PdCl2-4? If so, then:
<span>Cathode: [PdCl4]^-2(aq) + 2e- ======⇒ Pd(s) + 4Cl-(aq) </span>
<span>Anode: Cd(s) ==⇒ Cd+2(aq) + 2e-</span>
Well I’m. Or to sure but it can’t be B because when you throw the ball the the kinetic energy is still increasing