1) A 75.0-mL volume of 0.200 M NH3 ( K b=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 23.0 mL of HNO3. 2)A 52.0-mL volume of 0.35 M CH3COOH ( K a=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 21.0 mL of NaOH. Calculate the pH value of \\[{\\text{HN}}{{\\text{O}}_3}\\] solution containing \\[0.315{\\text{ g}}\\] acid in \\[{\\text{200 ml}}\\] of solution.\\[\\left( {{\\text
Calculate the pH of 0.100 L of a buffer solution that is 0.20 M in HF (Ka = 3.5 x 10-4 ) and 0.47 M in NaF. Express your answer using three significant figures. pH = 3.83 Part B What is the pH after adding 0.003 mol of HNO3 to the buffer described in Part A? Express your answer using three significant figures.

Calculate pH of HNO 3 using the pH equation. Because HNO 3 is a strong acid, it dissociates completely to H + and NO 3-ions in the water. HNO 3 → H + + NO 3-According to the stoichiometric ratios, H + concentration equals to the HNO 3 concentration. Therefore, we can directly substitute HNO 3 concentration to pH calculation equation. pH

Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 M HNO 3 (aq) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 40.0 mL.
Calculate the pOH by plugging the [OH −] into the equation. Calculate the pH by rearranging and plugging in the pOH: pH = 14.00 - pOH. Calculate the answers. pH = − log[H3O +] = − log(0.10) = 1.00. [OH −] = 0.0044M × 2 = 0.0088 M. pOH = − log[OH −] = − log(0.0088) = 2.06. pH = 14.00 − pOH = 14.00 − 2.06 = 11.94. Calculate the pH from the equation: pH = -log[H +] = -log[0.001] = 3. Then use the formula which shows the relationship between pH and pOH: pH + pOH = 14. and calculate pOH as: 14 - pH = 14 - 3 = 11
1) The equivalence point of an acid-base reaction (the point at which the amounts of acid and of base are just sufficient to cause complete neutralization). 2) The pH of the solution at equivalence point is dependent on the strength of the acid and strength of the base used in the titration.

pH = −log[2.3 ×10−5] = 4.64 (10.5.1) (10.5.1) pH = − log [ 2.3 × 10 − 5] = 4.64. When the pH of a solution is known, the concentration of the hydrogen ion can be calculated. The inverse of the logarithm (or antilog) is the 10x 10 x key on a calculator. [H+] = 10−pH (10.5.2) (10.5.2) [ H +] = 10 − pH.

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    • calculate ph of hno3