solving for pH and concentration A 1L flask has 2x10^-2 M HCl mixed with 2X10^-2 NaOH. Solve for pH and concentration?

[匿名]

2016-06-15 12:27 pm

So i started with balancing the equations and found that the species in equilibrium would be [H3O+],[Na+],[OH-],[Cl-]
I used the Kw = 10^-14 = [H3O+][OH-]
I rearranged this equation to be [OH-]=Kw/[H3O+] because I saw that in class but I am not sure as to why I am doing that.
and I already know that [Na+] and [Cl-] have 2x10^-2 (concentrations? am I right?)
but then i am not sure of the next steps.
Any pointers or explanations would be greatly appreciated.

回答 (1)

fooks

2016-06-15 12:51 pm

HCl + NaOH → NaCl + H₂O
Mole ratio HCl : NaOH = 1 : 1

You have not stated the volumes of the HCl and NaOH used. If the volumes are equal, the numbers of moles of moles HCl and NaOH are equal. Under such circumstances, both HCl and NaOH completely react, and the resulting solution is NaCl solution.

Na⁺ and Cl⁻ ions do not react with water, and thus they exhibit no effect on the pH of the solution. To determine the pH of the final solution, only the self-dissociation of water is needed to be considered.
2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

Kw = [H₃O⁺][OH⁻]
As [H₃O⁺] = [OH⁻], Kw = [H₃O⁺]²
1.0 × 10⁻¹⁴ = [H₃O⁺]²
Thus, [H₃O⁺] = 1.0 × 10⁻⁷ M

pH = -log[H₃O⁺] = -log(1.0 × 10⁻⁷) = 7.0

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