The acidic stream supplies approximately.01 M H⁺; the alkaline stream supplies10⁻⁶ M OH⁻ under the stipulated temperature-independent water-ion convention. Neutralize before volume averaging: $[H^+]\approx[2(.01)-3(10^{-6})]/5=.0039994$ M. Therefore $pH=-\log_{10}(.0039994)=2.3980$,rounded **2.40**. Arithmetic averaging of pH values is invalid because pH is logarithmic.
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