The purpose of the current experiment was to determine the pH of various hydrochloric acid and acetic acid solutions, to determine the pH of various salt solutions, to prepare a buffer solution, and determine the effects of adding a strong acid and strong base to the buffer solution versus adding a strong acid and strong base to water. The measured pHs for the hydrochloric acid solutions were 1.6, 2.2, 2.9, and 3.8. The measured pHs for the acetic acid solutions were 2.9, 3.9, 4.2, and 4.4. The pHs measured for the salts were 4.3 for sodium chloride, 7.3 for sodium acetate, 8.9 for sodium bicarbonate, 10.8 for sodium carbonate, 7.9 for ammonium chloride, and 6.9 for ammonium acetate. The pH for the buffer solution and the strong acid was 3.9 and the pH for the buffer solution and the strong base was 11.6. The pH of water with the strong acid added was 3.44 and the pH of water with the strong base added was 13.4
Introduction:
Water is a polar molecule, meaning many compounds readily dissolve in it forming aqueous solutions. The water molecules surround the molecules or ions in the solution for dissolution. In other cases, the water molecules react with the solute molecules or solute ions. In pure water, a small amount of molecules take part in an equilibrium reaction1. 2 H2O (l) H3O+ (aq) + OH- (aq)
The Essay on Strong Acid Solution Khp Point
Introduction The two types of aqueous solutions in chemistry that are being experimented with are called acids and bases. A strong acid solution is one that consists of a solution that is less than seven ph factor. The basic solution is one that consists of greater than seven on the ph scale. In this experiment standardization of a chemical reaction is used to determine the concentration of the ...
This self-ionization of water produces a small amount of both hydronium and hydroxide ions. The equilibrium constant expression for this reaction is called Kw (1.0×10-14), or the ion product of water1. From this number, if the concentration of hydronium is known the concentration of hydroxide can be found or vice versa. Strong acids are substances that ionize completely in water and produce hydronium ions. Therefore, if a strong acid is added to a solution, the concentration of hydronium is equal to the original concentration of acid added to the solution. A strong base is a substance that ionizes completely in water and produces hydroxide ions.
If a strong base is added to a solution, the concentration of hydroxide is equal to the concentration of strong base added to the solution. In aqueous solutions hydronium and hydroxide can have widely different values, but their product must always equal Kw1. If the hydroxide or hydronium concentration is known, the pH of a solution can be easily calculated by taking the negative log of the hydronium concentration. If pH of a solution is known, ten to the negative power of that pH will produce the hydronium concentration in the solution. pH = -log [H3O+]
10-pH = [H3O+]
The same rule applies for the concentration of hydroxide ions. If the concentration of hydroxide is known, one can determine the pOH of the solution. Likewise, if the pOH of the solution is known, one can determine the hydroxide concentration of the solution1. pOH = -log[OH-]
10-pOH = [OH-]
For weak acids, only 10% or less of the molecules dissociate and release hydronium. In solutions with weak acids, the concentration of hydronium is much less than the concentration of the weak acid. In a weak acid solution, the relationship of the concentration of hydronium to the acid equilibrium constant expression, Ka, for the acid can be defined by an equation:
Where [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid added to the solution. In this expression, the concentrations of the products and reactants are raised to the power equal to their coefficient in their respective balanced chemical equation. Using the equation of Ka, the concentration of hydroxide and thus the pH can be calculated. For the chemical equation:
The Term Paper on Acid-Base Titration Chemistry Formal Lab Writeup By A.Mm
... titrant added to the analyte to produce a final solution with a specific Hydronium concentration: [ H3O+] = (Vanalyte x Manalyte ) – (Vtitrant)(Mtitrant) / (Vtitrant + Vanalyte) ... and NaOH (Strong-Acid + Strong-Base) Trial 2 – CH3COOH and NaOH (Weak-Acid + Strong-Base) Trial 3 – NH3 and HCl (Weak-Base + Strong-Acid) (*) marks equivalence ...
HClO(aq) + H2O(l) H30+(aq) + ClO-(aq)
The Ka expression would be:
Where the Ka can be found in table 1 of the lab manual. From this equation, if the only unknown is the concentration of hydronium, it can be arranged to solve for the unknown concentration. Once the concentration of hydronium is known, the negative log of that is the pH for the solution. Weak bases use Kb to represent the base equilibrium constant expression1. When an ionic solution dissolves in water, it dissociates into a cation and anion. The ions are surrounded by water molecules and under this circumstance some salts will react with the water molecules through a process called hydrolsis1. The anions of strong acids and the cations of strong bases do not hydrolyze.
Therefore, the resulting pH is 7.01. Depending on whether a weak acid or base was added to the water, salts of different pHs can be formed. Basic salts are a product of the reaction between a strong base and a weak acid. Acidic salts are the product of the reaction between a strong acid and weak base1. There are also neutral salts which are the product of the reaction between a strong acid and strong base and salts that can be acidic, basic, or neutral because of the reaction between a weak acid and a weak base1. 3.0×10-8= QUOTE
Solutions that undergo a minimal pH change when an acid or base is added are called buffer solutions1. A buffer solution can be prepared by mixing a weak acid and a salt containing the anion of that acid, like HClO and NaClO. A buffer solution can also be prepared by mixing a weak base with a salt containing the cation of the base. The pH of a buffer solution can be calculated by using the Ka expression and substituting the buffer solution conditions (concentration of solutions in buffer)1. pH = -log (3.0×10-8) = 7.52
