Explanation:
If the concentration of a substance is changed, the equilibrium will shift to minimise the effect of that change. If the concentration of a reactant is increased the equilibrium will shift in the direction of the reaction that uses the reactants, so that the reactant concentration decreases. For example, decreased volume and therefore increased concentration of both reactants and products for the following reaction at equilibrium will shift the system toward more products.
<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
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<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
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<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
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<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Answer:
Explanation:
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40.0mL(1 L/1000 mL) = 0.040 L
<span>then plug into the formula M = moles/liters </span>
<span>0.035 M = moles/ 0.040L </span>
<span>multipy both sides by 0.040L, and you get 0.0014 moles </span>
<span>so the answer is 1</span>
The answer to that question is c