Question

What is the pH of when 0.047 L of 0.0082 M HCN is titrated to its equivalence point with 0.085 L of NaOH? The K_a of HCN is 4.9×10^–10.

Answer 1

HCN: Molarity = 0.0082 M ka=4.9x10-10 volume 0.0472 pką-logka Men Ipka_ -log (4.9x1000) no ob moles a 0.0082 X 0.047 20.000 2854 Ipka= 9.31 at equivalence point moles of moles of 0.000 3854 HON = Nash - volume of Nach = 0.0852 HON + Nagh - NaCN t to weak acid strong base salt I 0.000 38540.000 3854 o +0.000 3854 1-0.000 3854-0.000 3854 0.000 3854 Total volume =0.047+ 0.085 = 0.1322 n 0.0003854016029197 M=[] {NacN] = > 0.132 for this type of salt pH formula pH- + 2 pk w tp ka + jog c ] pkwal4 pH-t{14+ 9-31 +99 C0-00 29197)] pH = 10.388

pH formula's for salt's weak acid & strong base -salftito pH = + [pkw tpka + loge] strongacid & weakbase salt toho pt = ₃ [ pk w _pkke_loge]