First we need to know the pH of the tea... it can vary greatly but generally it's somewhere around 6.3 or 6.4. Since this is acidic, that means the hydronium ion concentration $$[H_3O^+]$$ is greater than the hydroxide ion concentration $$[OH^-]$$. The approximate concentrations agree with this conclusion:$$pH=-\log[H_3O^+]\\6.3=-\log[H_3O^+]\\10^{-6.3}=[H_3O^+]\\\ \\pOH=-\log[OH^-]\\pH+pOH=14\\6.3-\log[OH^-]=14\\\log[OH^-]=-7.7\\\left[OH^-\right]=10^{-7.7}$$