You will prepare 5 cuvettes with different concentration of sodium phosphate (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 0.05mM. Fill the table below (before coming to lab) and use it to prepare your cuvettes. Your blank for each reaction is the PNPP mixed with Tris-HCL and Pi. DO NOT ADD ENZYME AT THIS POINT!!! 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   0.05     100 1500 3.33   0.05     100 1500 20   0.05     100 1500 50   0.05     100 1500 100   0.05     100 1500

Question is in the image, I got 3.07E-8 mmol protons produced. Need to confirm if this is correct.

In one experiment, if 25 mL of sodium hydroxide (0.5M) is titrated with nitric acid solution giving the final reading as in the following table upon reaching the end point: i. Calculate the number of moles of OH– ii. Calculate the concentration of HNO3.

Help! Please solve Q5 all parts

5. What would be the pH of a 100.0 mL solution containing 0.24 M formic acid (HCHO2; Ka = 1.8x10-4) and 0.24 M sodium formate (NaCHO₂)?

Upon the addition of 30.00 mL of 0.350 M HCl to a 80.00-mL of the weak base amine (Kb = 4.95 x 10-10) whose concentration is 0.530M, the resulting pH is then equal to 5.20. True or False?

what is the pH of 9ml distilled water after adding 0.5 mL of 0.1 M HCL? I forgot to record the ph of this during lab, so a good estimate is appreciated.   (Please type answer no write by hend)

Only typed solution

A group of students perform a titration experiment. The buret tube is filled to the 0.00ml line with 0.100M NaOH solution. The E. flask contains 50.0ml ov 0.100M HC2H3O2 solution. The Ka of the acetic acid is: Ka = 1.8 x 10-5. A) Write the ionic equation for the conjugate acid reactihng with water. One product is the OH- polyatomic ion.  B) What is the pH of the equivalence point?

Determine the pH of a solution that is prepped with 0.25 M CH3NH3Cl and 0.15 M CH3NH2  (Kb = 3.7 x 10-4) Thank you!