A very weak acid has à Ka= 4.37 x M? (-) What is the [H+] of a solution of this acid that is 2.65

80mL of a 0.3M solution of hexapeptide Leu-His-Cys-Glu-Asn-Arg is adjusted to pH=pl. The solution is then titrated with 0.2M HCI to a final pH of 2.1. Sketch the titration curve, labelling the pH and volume axes. Indicate the volume of HCl needed to reach each relevant pKa value and equivalence point(s). Relevant pka values are: 2.1, 4.3, 6.0, 8.3, 9.8, and 12.5.

A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) = H+ (aq) + A¯(aq) The equilibrium concentrations of the reactants and products are [HA] = 0.220 M, [H+] = 3.00 × 10−4 M, and [A¯] = 3.00 × 10−4 M. Calculate the value of pKa for the acid HA. pKa =

Quinine ( C20 H24 N2 O2) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, pK, = 5.1 and pK, = 9.7 ( pKp = – log Kp). Only 1 g quinine will dissolve in 1920.0 mL of solution. Calculate the pH of a saturated aqueous solution of quinine. Consider only the reaction | Q+ H2O= QH+ + OH- described by pK, where Q = quinine. pH =

What is the pH of a 0.150 M solution of acetic acid (pK₂ = 4.75) at 25 °C?

The pOH of a basic solution is 3.75. What is [H⁺]?

pH of solution 14.00 12.00 10.00 8.00 6.00 First 4.00 equivalence point 2.00 0 First midpoint Second equivalence point Third midpoint pH = pKa=12.32 Third equivalence point HPO4(aq) + OH(aq) PO4(aq) + H2O(1) Second midpoint pH = pKa = 7.21 H2PO4 (aq) + OH(aq) HPO42 (aq) + H2O(1) Using the Henderson- Hasselback equation, show how to create 2L of a 0.1 M KPhos pH 7.5 buffer using K2HPO4 and KH2PO4. The chart to the left should help you understand what pKa to start with. Show your work. pH = pKa = 2.16 H3PO4(aq) + OH(aq) H2PO4(aq) + H2O(l) 25.0 50.0 75.0 100.0 Volume of NaOH added (mL)

Example   The reference range for blood pH is 7.35 – 7.42.  What is this range expressed as [H+] in nmol l-1?   Calculation:     pH   =  - log [H+]       7.35  =  - log [H+]       [H+]  =  antilog  -7.35      =  4.47 x 10-8 mol.l-1  =  44.7nmol.l-1            similarly:     pH 7.42 = 38.0 nmol.l-1       Range [H+] = 38 – 44.7 nmol.l-1   If the blood pH decreases in an acidosis from 7.42 to 7.15, what is the change in [H+] in nmol.l-1?       so this is example on the worksheet, but i still understand how to answer the question.