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chap-17-2011W-a

# 100m 0100m 0020m 0100m 0020m 0120 xm ph pk a

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Unformatted text preview: 0M 0M xM + - A (aq) 0.100M 0.100M 0.020M 0.100M + 0.020M (0.120 + x)M pH = pK a + log[0.100] [0.150] =3.68 pH = pK a + log[A - ] [HA] = − log(1.4 ×10-4 ) + log(0.12 0.13) = 3.854 − 0.035(Keep more digits during calculation) =3.82(Round to significant digits at the end) Addition of strong acid or base to Buffers What What is the pH after the addition of 0.020 mole HCl to after 0.020 1.00 L of the solution containing 0.150 M lactic acid 0.150 and 0.100 M sodium lactate? lactic acid Ka= 1.4 × 10-4 0.100 HA(aq) C’ 0.150 M Ci (0.150 + 0.020)M 0.020)M Cf (0.170- x)M H + (aq) 0M 0M xM + A - (aq) 0.100M (0.100 (0.100 - 0.020)M (0.080 + x)M pH = pK a + log[A - ] [HA] = − log(1.4 ×10-4 ) + log(0.080 0.170) = 3.854 − 0.3274 = 3.53 pH = 3.53 (very close to 3.68) (very Buffer Buffer Solutions Could a weak base and its conjugate acid be used weak...
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