Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Questions: 1· Write the balanced net ionic equations for the two acid-base react

ID: 566452 • Letter: Q

Question

Questions: 1· Write the balanced net ionic equations for the two acid-base reactions in this experiment. Sodium hydroxide and hydrochloric acid Sodium hydroxide and acetic acid 2. Titration experiments can have many different variations. Calculate the experimental results of the two following variations of a titration experiment. a. In one experiment, 0.606 M H:S04 was used to standardize a sodium hydroxide solution. 14.6 mL of the NaOH solution was required to neutralize 13.7 mL of the sulfuric acid solution. What was the molarity of the NaOH solution? Show your work. b. In a second experiment, 1.07 g of an unknown solid monoprotic acid is titrated with sodium hydroxide solution. It takes 11.68 mL of 0.582 M NaOH to titrate the acid sample. What is the molar mass of the acid? Show your work. When performing an acid-base titration, it is important to use an indicator that changes color close to the pH where the reaction is complete. HCI and NaOH perfectly neutralize each other at pH 7.0. For that titration we use phenolphthalein, which changes color near pH 7.0 Imagine that you are standardizing a NaOH solution with HCI, and you accidentally use an indicator that changes color at the acidic pH of 2.0-more acidic than lemon juice, but less acidic than the HCI solution you used. What effect would this mistake have on your calculated NaOH concentration? Would the calculated NaOH molarity be higher, lower, or the same as the actual molarity? Explairn why, using complete sentences. 4.

Explanation / Answer

Ans 1 :

The well balanced complete reaction for HCl and NaOh is given as :

HCl (aq) + NaOH (aq) = NaCl (s) + H2O (l)

The net ionic equation is given as :

H+ (aq) + OH- (aq) = H2O (l)

The well balanced complete reaction for NaOH and acetic acid

NaOH (aq) + CH3COOH (aq) = CH3COONa (s) + H2O (l)

The net ionic equation will be given as :

CH3COOH (aq) + OH- (aq) = CH3COO- (aq) + H2O (l)