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pOH = -log4×10^-1
= 1-log 4
ph= 14-1-log4
= 13-log 4
HCL->H +cl
pH=-log5×10^-1
=1-log 5
sesudah dicampul
nNH4OH =0,4×600=240mmol
nHCL = 0,5×400= 200mmol
NH4OH + HCL --> NH4cl + H2O
240. 200.
-200. -200. 200
40mmol. 200mmol
h+ = 10^-5 × 40/200
h+ = 2×10^-6
pH = -log 2×10^-6
pH = 6-log2