Question

Calculate the pH at the equivalence point in titrating 0.028 M solutions of each of the following with 0.031 M NaOH.

(a) nitric acid (HNO₃)

pH = ______

(b) benzoic acid (HC₇H₅O₂), K_a = 6.3e-05

pH = _______

(c) hypobromous acid (HBrO), K_a = 2.5e-09

pH = _______

Answer 1

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Al equivalence-paint malu of NQDK = mcakes UNDE M = M2/2 _ 0.025 x = 0.03 M XV HNO3 + NAON NaNO3 HD Both are strong ie, strong base s strong acid pH=7) NAOM b)_bumal of benzoic acid smmol of PASO 0 Volumcis not-givn_bwe Corrales considered text.com each sofn. bomol of benzoic acid = 0.025 M X 1 = 0.028 mole mol of Naon = 0.031 *V, > 0.63 Nmole so_0:028 meixib dogi mo! KVEL VIL) = 0.029 - 0.90L 0.031 I C E I Benzoic acid * Naon Sodiumbenzoate +420 0.029 mote - 0.035 mol -0.028 -0.08 +0.02% - 0.02% mol The solution get further hydrolyse to form basic solution [Las cooNa] = 0.029 mot = 0.025my. 1L +O-SOL 302 from pH = { pkw + pka + logg - ** (14+ 4.20 +log 0.0142) - (pH= 8:18 -pka = -log ke pke 796-3x10-5 -3 4.20 Scanned with CamScanner

عملهوتبرع له طقم more of_Hibroz mole of NAOH 0.028 m XIL = 0.031 Mx VCS) V(L) = 0.02%,= 0,90 L- 01031 + - HBoo L I (mule) 0.028 - Noon 0.028 -0.028 Nabro - +420 - -0.028 +0.028 0.028 [ NoBro] = 0.226 - 0.0142 m l·gol It forms acidic solution Pka = -log 25x109 38.60 pH = 2 (pkw + pka +log) =*414 + 8.60 +log 0.0442) = (10.38