You are planning to run a gel using a Tris-HCl buffer. You have 0.75 M Tris-HCl, pH 7.30 buffer, but you need your final buffer to be at a concentration of 100 mM and you need a total volume of 1.5 liter. What do you do? (Remember C1 V1 = C2 V2). How much water needs to be added, too?

Your goal is to make a buffer with a pH of 4.75 from acetic acid and sodium acetate.  Assume the pKa for acetic acid is 4.74.  What is the volume of 1 M sodium acetate added to make your buffer? Report your answer to the tenths place.   To do this, we use the Henderson-Hasselbach Equation pH = pkA + log (base/acid) where pkA = 4.74 for acetic acid,  pH = target pH and [base] /[acid] = x/(1-x)  since we don't know the concentration of the base, we will call it x  the acid is just 1-x, assuming the two add  up to 100% solve for x, and then you know the mL of base and acid to add.

CONSIDER THE LACTIC ACD BUFFER WITH LACTIC ACID), CH₂CH(OH) COOH AND ITS SALT SODIUM LACTATE, CH₂CH(OH)COo Na t ++₂0= "ott OH OH a) WHAT IS THE PURPOSE OF A BUFFER SYSTEM? 6) WHY IS THE SALT OF THE ACID NEEDED? How DOES THE BUFFER D) HOW DOES THE BUFFER REACT WHEN ACIT) (#30t) IS ADDED: REACT WHEN BASE (OH-) IS ADDED

What is the pH of a buffer system that is 0.25 M in NH3 and 0.26 M in NHẠC?? Кь NH3 %3D 1.8 х 10:5 NOTE: Even though this system involves the weak base NH3 and its conjugate acid NH4*, it is still possible to use the Henderson-Hasselbalch equation using pKą for the weak acid NH4* and the ratio of [conjugate base]/[conjugate acid]. You do not need to convert the equation and use a pКь value.

We add 10mL of 0.1M NaOH to 100mL of 0.1M CH3COOH/ 0.1M CH3COO^-. (Ka = 1.8 x10 ^-5). Using the Henderson-Hasselbach equation, pH=pKa + log [A-]/[HA] a) before the addition of NaOH b) after the addition of NaOH

Hi, I need help with C (H+ in half-neutralized solution), Ka for unknown acid, and part D-change in PH from part C.   Thank you so much!

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3. If you didn't have a pH meter, how would you make 1.00 L of 40.0 mM potassium phosphate buffer at pH 6.96, given ONLY potassium phosphate monobasic (KH₂PO4) and potassium phosphate dibasic (K₂HPO4)? These are the only two chemical available -no strong acid or base are needed. Include details such as masses and when/how you would bring the volume up to 1L. Show your calculations. Use pka2=7.20 for phosphoric acid.

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)?

pH - 10.5 pOH - 3.3 Equilibrium [OH–] -  Equilibrium [Mg2+] -  Molar Solubility -  Ksp -  I Need Help Finding Out The Rest Please !