Experimental Procedure, Part F. A 1.0- mL volume of 0.010 M H2SO3 is added to a mixture of 12 drops of 0.01 M HIO3, 8 drops of deionized water, and 1 drop of starch solution. A color change in the reaction mixture occured in 40 seconds.
a. Assuming 25 drops per milliliter, determine the initial molar concentration of IO3- after the mixing but before any reaction occurs (at time = 0).
b. At the same point in time, what is the molar concentration of H2SO3?
c. The rate of the reaction is measured by the disappearance of H2SO3. For the reaction mixture in this question, what is the reaction rate? Express the reaction rate with appropriate units.
a.
25 drops/mL means the 1.0 mL sample is 25 drops. Total drops = 25 + 12 + 8 + 1 = 46
concn HIO3 = 0.01 x (12 drops/46 drops) = ??
b.
concn H2SO3 = 0.01 x (25 drops/46 drops) = ??
c.
rate rxn = 1/5 rate disappearance H2SO3. You need a balanced equation for this and it is as follows but check me out on this.
2HIO3 + 5H2SO3 ==> I2 + 5H2SO4 + H2O
a. Well, if we assume that there are 25 drops per milliliter, then we can calculate the initial molar concentration of IO3- after the mixing but before any reaction occurs (at time = 0). Since there are 12 drops of 0.01 M HIO3, and each drop is approximately 1/25th of a milliliter, we can calculate the initial molar concentration as follows:
Molar concentration of IO3- = (12 drops * 0.01 M) / (25 drops/mL)
b. Now, let's shift our attention to the molar concentration of H2SO3 at the same point in time. The experimental procedure mentions that a 1.0-mL volume of 0.010 M H2SO3 is added. Therefore, the molar concentration of H2SO3 remains the same as the initial concentration, which is 0.010 M.
c. The rate of the reaction is determined by the disappearance of H2SO3. Since the color change in the reaction mixture occurs in 40 seconds, we can express the reaction rate as the change in concentration of H2SO3 per unit time. However, to provide the appropriate units, we need to know the volume of the reaction mixture. Could you provide that information?
a. To determine the initial molar concentration of IO3- after mixing but before any reaction occurs, we need to consider the number of drops of HIO3 and the volume of the mixture.
Given:
- 12 drops of HIO3
- 25 drops per milliliter
We can calculate the volume of HIO3 in milliliters:
Volume of HIO3 = (12 drops) / (25 drops/mL)
Volume of HIO3 = 0.48 mL
The initial molar concentration of IO3- is given as 0.01 M. Since IO3- is the only species in the reaction with a molar concentration, the concentration of IO3- after mixing but before any reaction occurs is also 0.01 M.
Answer: The initial molar concentration of IO3- after mixing but before any reaction occurs is 0.01 M.
b. The volume of H2SO3 is given as 1.0 mL, and the initial molar concentration of H2SO3 is given as 0.010 M. Therefore, the molar concentration of H2SO3 after mixing but before any reaction occurs is also 0.010 M.
Answer: The molar concentration of H2SO3 after mixing but before any reaction occurs is 0.010 M.
c. The rate of the reaction is measured by the disappearance of H2SO3. Since the color change occurred in 40 seconds, we can use this information to calculate the reaction rate.
The reaction rate is defined as the change in concentration of a reactant or product over time. In this case, we are measuring the disappearance of H2SO3. The change in concentration of H2SO3 can be calculated using the initial and final concentrations of H2SO3 and the time taken for the reaction.
Given:
- Initial concentration of H2SO3 = 0.010 M
- Final concentration of H2SO3 = 0 M (H2SO3 completely reacts)
- Time = 40 seconds
The change in concentration of H2SO3 can be calculated as:
Change in concentration of H2SO3 = (Final concentration of H2SO3) - (Initial concentration of H2SO3)
Change in concentration of H2SO3 = 0 M - 0.010 M
Change in concentration of H2SO3 = -0.010 M
Since the rate of the reaction is negative (due to the decrease in concentration of H2SO3), we take the absolute value and divide it by the time:
Reaction rate = |Change in concentration of H2SO3| / Time
Reaction rate = |-0.010 M| / 40 seconds
Reaction rate = 0.010 M / 40 seconds
Reaction rate = 0.00025 M/s
Answer: The reaction rate for the disappearance of H2SO3 in the reaction mixture is 0.00025 M/s.
a. To determine the initial molar concentration of IO3-, we need to convert the given volume and concentration of HIO3 drops into milliliters.
1. Start by converting the number of drops of HIO3 into milliliters. Since there are 25 drops per milliliter, we divide the number of drops by 25:
12 drops ÷ 25 drops/mL = 0.48 mL
2. Next, we convert the volume of HIO3 from milliliters to liters since the concentration is given in moles per liter (M):
0.48 mL = 0.48 mL × (1 L / 1000 mL) = 0.00048 L
3. Now, with the volume and concentration of HIO3, we can calculate the initial moles of IO3-:
Moles of IO3- = Volume (L) × Concentration (M)
Moles of IO3- = 0.00048 L × 0.01 M = 4.8 × 10^-6 moles
Thus, the initial molar concentration of IO3- after mixing but before any reaction occurs (at time = 0) is 4.8 × 10^-6 M.
b. The molar concentration of H2SO3 can be determined by assuming the volume of the solution remains constant, even after adding drops of HIO3, water, and starch solution. Thus, the initial molar concentration of H2SO3 remains the same as the given concentration, which is 0.010 M.
c. The reaction rate is determined by the disappearance of H2SO3 over time. Here, it is mentioned that the color change in the reaction mixture occurs in 40 seconds. From this information, we know that the reaction is complete within 40 seconds.
Since the concentration of H2SO3 is not provided, we cannot explicitly calculate the reaction rate. However, the reaction rate can be expressed in terms of the change in concentration of H2SO3 per unit time, such as moles per liter-second (M/s) or moles per milliliter-second (M/mL-s), given the appropriate measurements.