Weak Acid Titration Report
Weak Acid Titration
Abstract: Our method for determining the unknown weak acid was to determine the equilibrium constant K from the molecular weight of the weak acid from our titration data. In this lab the acid Potassium hydrogen phthalate and two unknown acids were titrated. We determined the molar mass of the Potassium hydrogen phthalate, for the unknown acids we calculated the molar mass and the Ka values. We used NaOH as the known base for titrating in all three of the titrations. Our ka value based on our titration curve of KHP was 6.9 (the equivalence point where the pH equals the pKa) and our unknown acids were determined to be Hexanoic Acid (with an actual molar mass of 116.6g) with a determined mass of 116.1g. The unknown A3 …show more content…
It is also a cheap and effective way of doing so. One disadvantage is that is a slow process and there are many potential sources for error such as an over titration. It is accurate when done correctly but very easy to commit an error when done incorrectly. ). For any given acid reaction in water the acid donates a proton to a water molecule producing a hydronium ion (HA+H2OH3O++A-). At a given temperature, the concentrations of this equilibrium can be expressed by the weak acid dissociation constant (Ka=[H3O+]eq*[A-]eq/[HA]eq). In order to integrate a determinable factor into this equation, pH, it may be rearranged to solve for the hydronium ion concentration ([H3O+]eq=Ka*[HA]eq/[A-]eq) and then, as pH is equal to the negative logarithm of the hydronium ion concentration, taking the negative logarithm of both sides of the equation (-log[H3O+]eq=-log(Ka)+log([A-]eq/[HA]eq)). This can then be expressed in terms of pH and pKa (pH=pKa+log([A-]eq/[HA]eq)). Due to the fact that at the equilibrium point the reactants are completely converted to products, it would stand to reason that at the point in the reaction halfway to the equilibrium point that there would be equal concentrations of product ([A-]) and reactant ([HA]). Therefore, at this half equilibrium point the concentrations of product and reactant would divide to equal one for substitution into the pKa equation, and as the context of this
Abstract: Our method for determining the unknown weak acid was to determine the equilibrium constant K from the molecular weight of the weak acid from our titration data. In this lab the acid Potassium hydrogen phthalate and two unknown acids were titrated. We determined the molar mass of the Potassium hydrogen phthalate, for the unknown acids we calculated the molar mass and the Ka values. We used NaOH as the known base for titrating in all three of the titrations. Our ka value based on our titration curve of KHP was 6.9 (the equivalence point where the pH equals the pKa) and our unknown acids were determined to be Hexanoic Acid (with an actual molar mass of 116.6g) with a determined mass of 116.1g. The unknown A3 …show more content…
It is also a cheap and effective way of doing so. One disadvantage is that is a slow process and there are many potential sources for error such as an over titration. It is accurate when done correctly but very easy to commit an error when done incorrectly. ). For any given acid reaction in water the acid donates a proton to a water molecule producing a hydronium ion (HA+H2OH3O++A-). At a given temperature, the concentrations of this equilibrium can be expressed by the weak acid dissociation constant (Ka=[H3O+]eq*[A-]eq/[HA]eq). In order to integrate a determinable factor into this equation, pH, it may be rearranged to solve for the hydronium ion concentration ([H3O+]eq=Ka*[HA]eq/[A-]eq) and then, as pH is equal to the negative logarithm of the hydronium ion concentration, taking the negative logarithm of both sides of the equation (-log[H3O+]eq=-log(Ka)+log([A-]eq/[HA]eq)). This can then be expressed in terms of pH and pKa (pH=pKa+log([A-]eq/[HA]eq)). Due to the fact that at the equilibrium point the reactants are completely converted to products, it would stand to reason that at the point in the reaction halfway to the equilibrium point that there would be equal concentrations of product ([A-]) and reactant ([HA]). Therefore, at this half equilibrium point the concentrations of product and reactant would divide to equal one for substitution into the pKa equation, and as the context of this