Chemistry: water equilibrium question

Could you solve this with full working? Thanks! :)

When you suffer from an acid stomach the fluid in your stomach can contain HCl at a concentration of 2.50x10^-4 molL-1. To reduce this acid you are advised to take an antacid tablet that contains 5.00mg of Al(OH)3 and 5.0mg of Mg(OH)2. At this time you have 1.5L of fluid in your stomach, calculate the concentration of hydrogen ions and hydroxide ions in your stomach after you take this tablet.

First we calculate the number of moles HCl in your stomach.

you have 1.5L of 2.5*10^-4mol/L (corresponding to 2.5*10^-4 mole in 1L)
so in1.5L you have 2.5*10^-4*1.5=3.75*10^-4 moles HCl in your stomach

molar mass of Mg(OH)2 corresponds to 24.3+2*(16+1)=58.3g
and 5mg corresponds to 5*10^-3/58=8.58*10^-5 moles of Mg(OH)2
Mg (OH)2+HCl -->MgCl2+2H2O so 8.58*10^-5moles neutralize 1.72*10^-4M HCl
for Al(OH)3 you have Al(OH)3+3HCl =AlCl3+3H2O
molar mass of Al(OH)3= 27+3*17=78g and 5mg correspond to 5*10^-3/78=6.41*10^-5 moles
as Al(OH)3 is trivalent you neutralize 6.41*10^5*3=1.92*10^-4M HCl

so you neutralize (1.92+1.72)*10^-4=3.64*10^-4M HCL

it remains in your stomach (3.75-3.64)*10^-4=0.11*10^-4 moles in 1.5L
so molarity 0.11*10^-4/1.5=7.3*10^-6M ions [H+] pH=5.13

concentrations of ionS [OH-] given by [OH-]*[H+] =KH2O at 37C which is not 14!!!

if you have table of KH2O you calculate