$n_{H^{+}} = 0.01 mol $
$n_{OH^{-}} = 2\times 1.07\times 6\div 100\div 40 = 0.00321 mol$
                                  $H^{+}             +           OH^{-} \rightarrow                   H_{2}O$
   Trước P/ứ          0.01 mol                        0.00321 mol
    P/ứ                      0.00321 mol               0.00321 mol                             0.00321 mol 
    Sau P/ứ             0.00679 mol                     0

$[H^{+}] = 0.00679\div 0.102 = 0.067 M$
$\rightarrow pH = 1,2$