pay more attention in class. pH = pKa + log [A-]/[HA Im not helping you anymore
preparation of 5.8 ph phosphate buffer
na+oh +25=53 ph+poh=2.3
Yes. Some alkalis do have a pH above 9. e.g. 46% NaOH solution.
See the Related Links for "Columbia.edu: Phosphate buffer automatic calculator" to the bottom for the answer.
You prepare 0.05mM from NaHC03 by first dissolving it. As you dissolve it in water, monitor the pH of the solution. The Internet does not list anything with the formula ZnSO4.7H20.
to prepare 100ml of 100mM Trissolution: Mol wt of Tris=121.14121.14g in 1000ml ----> 1M12.11g in 100ml -------->1M1M=1000mM121.1g---->1000mM12.11g ----------->100mM1.211g in 100ml and 100mM Tris
6g Tris HCl + 100ml dH2O, pH 6.8
Prepared 0.5M succinic acid and 0.5 NaOH and adjusted pH to desired value
Decide on the concentration of the buffer, use 1L to be simple PH for your buffer should be within one pH unit from the pKa of the acid/conjugate base use Henderson Hasselbalch Equation pH = pKa + log ([Base]/[Acid]) For a 1 M buffer [Acid] + [Base] = 1
pH=10, so pOH=14-10=4 [OH]=10^-pOH [OH]=10^-4 [OH]=0.0001
Tartaric acid - 18.75 mM Sodium Tartrate Dihydrate - 6.25 mM pH of this mixture should be near 2.9, adjust pH down to 2.5 using HCl, and QS to 1 L by water.
dissolve the 12 g of crystals of sodium phosphate in water to make 1oo ml