Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O
Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O
Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L
∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = ![\frac{number of moles of [OH-]}{volume of solution (L)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bnumber%20of%20moles%20of%20%5BOH-%5D%7D%7Bvolume%20of%20solution%20%28L%29%7D%20)
=

= 0.03199 M
<span>Electronegativity is the property of an element that measures the
ability of it to attract and form electron bonds. The trend in the periodic
table in terms of electronegativity decreases from right to left and from top
to bottom. In the case of period 4, the element with the highest electronegativity
is bromine. </span>
Henderson–Hasselbalch equation is given as,
pH = pKa + log [A⁻] / [HA]
-------- (1)
Solution:
Convert Ka into pKa,
pKa = -log Ka
pKa = -log 1.37 × 10⁻⁴
pKa = 3.863
Putting value of pKa and pH in eq.1,
4.29 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.29 - 3.863
log [lactate] / [lactic acid] = 0.427
Taking Anti log,
[lactate] / [lactic acid]
= 2.673
Result:
2.673 M
lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.