The correct option is D 3.46
pHsolution1=3 ; C1=[H+]=10−3pOHsolution2=4 ; C2=[OH−]=10−4Cf=[H+]remains after neutralization V1=400 mL , V2=600 mLVf=V1+V2=1000 mLApplying:
C1V1−C2V2=CfVfCf=C1V1−C2V2Vf=(10−3× 400)−(10−4× 600)1000 =3.4×10−4pH=− log10[3.4×10−4]pH=3.46