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pH of 0.55 M Potassium Formate HCOOK
On the calculation of pH value of potassium formate solution
There is a\ (0.55M\) potassium formate\ (HCOOK\) solution, find its\ (pH\) value. Potassium formate is a strong base and weak acid salt. Weak acid radical ions\ (HCOO ^ -\) in water will occur hydrolysis reaction.
The formula for hydrolysis reaction is:\ (HCOO ^ - + H_2O\ rightleftharpoons HCOOH + OH ^ -\). Let the hydrolysis equilibrium,\ (OH ^ -\) ion concentration is\ (x mol/L\). At the beginning, the concentration of\ (HCOO ^ -\) is\ (0.55mol/L\), then at equilibrium, the concentration of\ (HCOO ^ -\) is\ ((0.55 - x) mol/L\), and the concentration of\ (HCOOH\) is\ (x mol/L\). The ionization constant of
formic acid\ (HCOOH\) \ (K_a\) is a given value. From the ionization constant of water\ (K_w = K_a\ times K_ {h}\) (\ (K_ {h}\) is the hydrolysis constant), the hydrolysis constant\ (K_ {h} =\ frac {K_w} {K_a}\) can be obtained.
For this hydrolysis reaction, the hydrolysis constant expression is\ (K_ {h} =\ frac {[HCOOH] [OH ^ -]} {[HCOO ^-]}\)， that is\ (\ frac {K_w} {K_a} =\ frac {x\ cdot x} {0.55 - x}\).
Because in general, the degree of hydrolysis of potassium formate is small, so\ (0.55 - x\ approx 0.55\), then\ (\ frac {K_w} {K_a} =\ frac {x ^ 2} {0.55}\).
The ionic product constant of water is known\ (K_w = 1.0\ times10 ^ {-14}\), and the ionization constant of formic acid\ (K_a\) can be found. Substitute the value into the above equation to obtain the value of\ (x\), that is, the concentration of\ (OH ^ -\) ions.
Then from\ (pH = 14 - pOH\),\ (pOH = -\ lg [OH ^ -]\), and then calculate the\ (pH\) value of\ (0.55M\) potassium formate solution.
After rigorous reasoning and calculation, the pH value of this solution can be accurately obtained, which is the application of chemical principles in practical problem solving.
There is a\ (0.55M\) potassium formate\ (HCOOK\) solution, find its\ (pH\) value. Potassium formate is a strong base and weak acid salt. Weak acid radical ions\ (HCOO ^ -\) in water will occur hydrolysis reaction.
The formula for hydrolysis reaction is:\ (HCOO ^ - + H_2O\ rightleftharpoons HCOOH + OH ^ -\). Let the hydrolysis equilibrium,\ (OH ^ -\) ion concentration is\ (x mol/L\). At the beginning, the concentration of\ (HCOO ^ -\) is\ (0.55mol/L\), then at equilibrium, the concentration of\ (HCOO ^ -\) is\ ((0.55 - x) mol/L\), and the concentration of\ (HCOOH\) is\ (x mol/L\). The ionization constant of
formic acid\ (HCOOH\) \ (K_a\) is a given value. From the ionization constant of water\ (K_w = K_a\ times K_ {h}\) (\ (K_ {h}\) is the hydrolysis constant), the hydrolysis constant\ (K_ {h} =\ frac {K_w} {K_a}\) can be obtained.
For this hydrolysis reaction, the hydrolysis constant expression is\ (K_ {h} =\ frac {[HCOOH] [OH ^ -]} {[HCOO ^-]}\)， that is\ (\ frac {K_w} {K_a} =\ frac {x\ cdot x} {0.55 - x}\).
Because in general, the degree of hydrolysis of potassium formate is small, so\ (0.55 - x\ approx 0.55\), then\ (\ frac {K_w} {K_a} =\ frac {x ^ 2} {0.55}\).
The ionic product constant of water is known\ (K_w = 1.0\ times10 ^ {-14}\), and the ionization constant of formic acid\ (K_a\) can be found. Substitute the value into the above equation to obtain the value of\ (x\), that is, the concentration of\ (OH ^ -\) ions.
Then from\ (pH = 14 - pOH\),\ (pOH = -\ lg [OH ^ -]\), and then calculate the\ (pH\) value of\ (0.55M\) potassium formate solution.
After rigorous reasoning and calculation, the pH value of this solution can be accurately obtained, which is the application of chemical principles in practical problem solving.
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