URGENT!!!!! 4 questions total!
1. .0025 M tartric acid completely reacts with .0030 m of al bicarbonate if .913 l of co2 forms at 298 k with pressure of 1.4 ATM how much base used?
2. 3.1 x 10^23 molec of .0035 m hydriodic acid reacts with 25 g of magnesium bicarbonate how much carbon dioxide if at stp?
3. 50.0 ml of lead (2) phosphate reacts with enough ammonium chloride. Following data collected. What is molarity of lead phosphate solution?
T1 1.56
T2 1.67
T3 1.51
4. What is ph and poh of solution made by adding 500 ml of water to 250 ml of 7.5 x 10^-4 M of NaOH?
1. To find out how much base was used, we need to calculate the amount of tartric acid that reacted with the bicarbonate. First, determine the moles of tartric acid:
Moles of tartric acid = concentration (M) x volume (L)
= 0.0025 M x 0.913 L
= 0.00228 moles of tartric acid
Since the reaction between tartric acid and bicarbonate is 1:1, the moles of base used will be the same as the moles of tartric acid:
Moles of base used = 0.00228 moles
2. To find out the amount of carbon dioxide produced, we need to calculate the moles of magnesium bicarbonate reacting with the hydriodic acid. First, calculate the moles of hydriodic acid:
Moles of hydriodic acid = concentration (M) x volume (L)
= 0.0035 M x (3.1 x 10^23 molecules / 6.022 x 10^23 molecules/mol)
= 0.00181 moles of hydriodic acid
The balanced chemical equation is:
2HI + Mg(HCO3)2 -> MgI2 + 2H2O + 2CO2
From the balanced equation, we can see that 1 mole of magnesium bicarbonate produces 2 moles of carbon dioxide. Therefore, we need to double the amount of moles of hydriodic acid to find the moles of carbon dioxide:
Moles of carbon dioxide = 2 x 0.00181 moles
= 0.00362 moles of carbon dioxide
Since the question states that the reaction is at STP (Standard Temperature and Pressure), we can use the molar volume of a gas at STP to find the volume of carbon dioxide:
Volume of carbon dioxide at STP = moles of carbon dioxide x molar volume (22.4 L/mol)
= 0.00362 moles x 22.4 L/mol
= 0.0811 L or 81.1 mL
Therefore, the amount of carbon dioxide produced is 81.1 mL.
3. To find the molarity of the lead(II) phosphate solution, we need to use the titration data provided. From the data, we can see that the volume of the lead(II) phosphate solution is not given, so we cannot directly determine the molarity using the given information. More information is needed to calculate the molarity.
4. To calculate the pH and pOH of the solution made by adding water to NaOH, we can use the equation:
pH + pOH = 14
First, calculate the concentration of NaOH in the solution:
Concentration of NaOH = 7.5 x 10^-4 M
Next, calculate the moles of NaOH:
Moles of NaOH = concentration (M) x volume (L)
= 7.5 x 10^-4 M x 0.250 L
= 1.88 x 10^-4 moles of NaOH
Since the volume of water (500 mL) is much larger than the volume of NaOH (250 mL), we can assume that the volume of the resulting solution remains at 500 mL. Therefore, the final concentration of NaOH will be:
Final concentration of NaOH = (moles of NaOH) / (final volume of solution in L)
= (1.88 x 10^-4 moles) / 0.500 L
= 3.76 x 10^-4 M
Now we can calculate the pOH:
pOH = -log10 (concentration of hydroxide ion)
= -log10 (3.76 x 10^-4)
≈ 3.42
Finally, use the equation pH + pOH = 14 to find the pH:
pH = 14 - pOH
= 14 - 3.42
≈ 10.58
Therefore, the pH of the solution is approximately 10.58, and the pOH is approximately 3.42.
1. To find out how much base was used, we need to determine the stoichiometry of the reaction between the tartric acid and aluminum bicarbonate. The balanced reaction equation is required to find the mole ratios. Once we know the mole ratio, we can use the given amount of tartric acid to calculate the amount of base used.
2. To calculate the amount of carbon dioxide produced, we also need the balanced reaction equation between hydriodic acid and magnesium bicarbonate. Using the stoichiometry, we can determine the mole ratio and then calculate the amount of carbon dioxide produced.
3. To determine the molarity of the lead (II) phosphate solution, we need to use the given data, which consists of three different measurements of time (T1, T2, and T3). These times correspond to the precipitation of lead (II) phosphate when reacted with ammonium chloride. The changing color or precipitation indicates the endpoint of the reaction against the known concentration of standard ammonium chloride. By analyzing the data, we can find the concentration of lead (II) phosphate.
4. To find the pH and pOH of the solution, we first need to calculate the concentration of hydroxide ions from the given concentration of NaOH. By using the concentration of hydroxide ions, we can determine the pOH of the solution. The pH can be found by subtracting the pOH from 14. The fact that water is added to the solution indicates that the concentration of hydroxide ions will change. However, the change will be small since the initial concentration of NaOH is low. By calculating the new concentration, we can then find the pH and pOH of the resulting solution.