solution pH acidic, basic or neutral NH4Cl     NaHCO3     KNO3

This is a Dry Lab since we cannot meet on Campus. After calibrating a pH meter you would have measured the pH of four different Basic Aqueous Solutions and examined the data for trends. You would have calculated the expected pH of each of the strong base solutions by assuming that for every NaOH formula unit that dissolves, one OH- ion is released. Thus, we can assume that the OH- Molarity equals the NaOH Molarity. Using the OH- Molarity and the water ionization equilibrium expression (Kw= 1.0 X 10 to the -14 power = H+ Molarity X OH- Molarity), we can calculate the H+ Molarity of the solution. Using the H+ Molarity, we can calculate the solution's pH. Thus a 1.0 X 10 to the 4 power NaOH solution has a pH of 10. The calculated pH value is arrived at as follows. By rearranging the water ionization equilibrium expression and plugging in the OH- Molarity, we can solve for H+ Molarity. H+ M = 1.0 X 10 to the -14 power divided by 1.0 X 10 to the -4 power 1.0 X 10 to the 10 power. pH= -log…

Prepare another 5 cuvettes with different concentration of Pi using the stock 1mM Pi, 0.5mM PNPP solution (in water) and 0.2M Tris-HCl, pH=8.00. For this part, the concentration of PNPP will remain constant at 0.150mM. Fill the table below (before coming to lab) and use it to prepare your cuvettes. The completed table must be in your pre-lab. Your blank for each reaction is the PNPP mixed with Tris-HCL and Pi. Again, prepare one cuvette at the time to minimize exposure of the PNPP to the air. You can prepare your next cuvette while your previous one is being read.  previous one is being read.  Final [Pi]uM Volume of 1mM Pi Final [PNPP] mM Volume of 0.5mM PNPP Volume of 0.2M Tris-HCl (uL) Volume of enzyme (uL) Total Volume (uL) 0   0.150     100 1500 3.33   0.150     100 1500 20   0.150     100 1500 50   0.150     100 1500 100   0.150     100 1500

#6 pH and pOH calculations 1. Determine the pH, pOH, [H], and [OH-] for each of the following problems. You may assume each acid and base listed will 100% dissociate. a. 8.50 x 10-³ M HBr b. 5.00 mL of 0.250M HCIO4 diluted to 50.0 mL (M₁V₁ = M₂V₂) c. 1.52g of HNO3 in 575 mL of solution. Be sure to convert to molarity. d. 1.50 x 10-³ M Sr(OH)2 solution e. 2.250 g of LiOH dissolved in 250.0 mL of solution. Be sure to convert to molarity.